Physics 1 - Problems With Solutions

Physics 1. Mechanics Problems with solutions

1

Coordinates and vectors.

Problem 1.1. coordinates). Solution.

Find the distance between two points P1 = (r, ϕ1 ) and P2 = (r, ϕ2 ) (polar

s2 = (x2 − x1 )2 + (y2 − y1 )2 = (r cos ϕ2 − r cos ϕ1 )2 + (r sin ϕ2 − r sin ϕ1 )2 = 2r2 (1 − cos ϕ1 cos ϕ2 − sin ϕ1 sin ϕ2 ) = 2r2 [1 − cos(ϕ2 − ϕ1 )]   ϕ2 − ϕ1 2 2 = 4r sin 2

Problem 1.2. coordinates). Solution.

Find the distance between P1 = (R1 , θ1 , ϕ1 ) and P2 = (R2 , θ2 , ϕ2 ) (spherical

s2 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 = (R2 sin θ2 cos ϕ2 − R1 sin θ1 cos ϕ1 )2 + (R2 sin θ2 sin ϕ2 − R1 sin θ1 sin ϕ1 )2 + (R2 cos θ2 − R1 cos θ1 )2 = R12 + R22 − 2R1 R2 cos θ1 cos θ2 − 2R1 R2 sin θ1 sin θ2 cos(ϕ2 − ϕ1 )

Problem 1.3. We define elliptical coordinates as follows: r = coordinates. Find x, y as functions of r, ϕ. Solution. y = x tan ϕ →

1

p x2 /a2 + y 2 /b2 and ϕ as in polar

Physics 1. Mechanics

Problems x2 x2 tan2 ϕ + = r2 → a2 b2 x = abr cos ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2 y = abr sin ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2

p Problem 1.4. In elliptical coordinates (r = x2 /a2 + y 2 /b2 and ϕ as in polar coordinates) find distance between two points P1 = (r, ϕ1 ) and P2 = (r, ϕ2 ). Solution. s2 = (x2 − x1 )2 + (y2 − y1 )2 = (abr)2 + (abr)2

cos ϕ2 p

a2 cos2 ϕ2 + b2 sin2 ϕ2 sin ϕ2

−p

cos ϕ1

!2

a2 cos2 ϕ1 + b2 sin2 ϕ1 sin ϕ1

!2

−p a2 cos2 ϕ2 + b2 sin2 ϕ2 a2 cos2 ϕ1 + b2 sin2 ϕ1 (abr)2 (abr)2 + = 2 a cos2 ϕ1 + b2 sin2 ϕ1 a2 cos2 ϕ2 + b2 sin2 ϕ2 (abr)2 cos(ϕ2 − ϕ1 ) p − 2p a2 cos2 ϕ1 + b2 sin2 ϕ1 a2 cos2 ϕ2 + b2 sin2 ϕ2 p

Problem 1.5. Two different coordinate systems are established on a straight line, x and x0 , which are related as follows: x/a = (x0 /b)3 , where a and b are constants. The distance element in terms of x is given by ds2 = dx2 . Find the expression for the distance element in terms of x0 . Solution. dx 2 2 ds2 = dx2 = ( 0 )2 dx0 = (3ax02 /b3 )2 dx0 dx Problem 1.6. Two different coordinate systems are established on a plane, (x, y) and (x0 , y 0 ), which are related as follows: x0 = x cos θ − y sin θ,

y 0 = x sin θ + y cos θ,

where θ = const. Find the distance element in terms of coordinates (x0 , y 0 ), if (x, y) are Cartesian. Solution. Inverting (solving for x and y) we get x = x0 cos θ + y 0 sin θ,

y = −x0 sin θ + y 0 cos θ

2

Physics 1. Mechanics

Problems

Now ds2 = dx2 + dy 2 = (

∂x 0 ∂x 0 2 ∂y 0 ∂y 0 2 dx + dy ) + ( dx + 0 dy ) ∂x0 ∂y 0 ∂x0 ∂y 2

= (cos θdx0 + sin θdy 0 )2 + (− sin θdx0 + cos θdy 0 )2 = dx0 + dy 0

Problem 1.7.

2

Same as above but the relation reads x0 = x cos θ + y sin θ,

y 0 = x sin θ + y cos θ,

. Solution. Inverting one gets x0 cos θ − y 0 sin θ x= cos2 θ − sin2 θ −x0 sin θ + y 0 cos θ y= cos2 θ − sin2 θ Now (cos2 θ − sin2 θ = cos 2θ) ds2 = dx2 + dy 2 = (

∂x 0 ∂x 0 2 ∂y ∂y dx + 0 dy ) + ( 0 dx0 + 0 dy 0 )2 0 ∂x ∂y ∂x ∂y

1 [(cos θdx0 − sin θdy 0 )2 + (− sin θdx0 + cos θdy 0 )2 ] cos2 2θ 1 2 2 [dx0 + dy 0 − 2 sin 2θdx0 dy 0 ] = cos2 2θ

=

Problem 1.8. New coordinates (ρ0 , ϕ0 ) are introduced which are related to the ordinary polar coordinates (ρ, ϕ) as follows: ϕ0 = ϕ, ρ0 = 1/ρ. Find the distance element. Solution. ϕ0 = ϕ, ρ0 = 1/ρ → ϕ = ϕ0 , ρ = 1/ρ0 ∂ρ ∂ρ 0 2 ∂ϕ 0 ∂ϕ 0 2 ds2 = dρ2 + ρ2 dϕ2 = ( 0 dρ0 + dϕ ) + (1/ρ0 )2 ( dρ + dϕ ) 0 ∂ρ ∂ϕ ∂dρ0 ∂ϕ0 2

= (1/ρ0 )4 dρ0 + (1/ρ0 )2 dϕ0

2

Problem 1.9. Starting with cartesian coordinates we define new ones as follows: x0 = ax + by, y 0 = cx + dy, where a, b, c, d are some constant parameters. What conditions on these parameters should be satisfied in order that the new coordinates also be cartesian and the measure of the distance remain the same ?

3

Physics 1. Mechanics

Problems

Solution. 2

2

ds2 = dx2 + dy 2 = dx0 + dy 0 → (adx + bdy)2 + (cdx + ddy)2 = dx2 + dy 2 → (a2 + b2 )dx2 + (c2 + d2 )dy 2 + 2(ab + cd)dxdy = dx2 + dy 2 → a2 + c2 = b2 + d2 = 1,

ab + cd = 0

d = −ab/c → b2 + (ab/c)2 = 1 → (b/c)2 = 1 → c = b or c = −b if c = b → d = −a if c = −b → d = a and a2 + b2 = 1

Problem 1.10. Find the volume. Solution.

Four vertices of the tetrahedron are given by four radius-vectors ri , i = 1, 2, 3, 4.

V = |(r4 − r1 ) × (r3 − r1 ) · (r2 − r1 )| Problem 1.11. Let r1 be the vector from the center of the Earth to New-York and r2 is the vector from the center of the Earth to Jerusalem. Find the angle between the two vectors (find the lacking data in a geographical atlas). p Problem 1.12. Find the distance element ds2 in the elliptical coordinates r = x2 /a2 + y 2 /b2 , tan ϕ = y/x. Define unit vectors tangential and normal to r = const curves and derive their relation to ex and ey . Problem 1.13. Are the coordinates x0 = xy, y 0 = y/x legitimate ? If yes, find the distance element. Solution. Not good: x, y and −x, −y have the same x0 , y 0 . Let us limit ourselves with x > 0, y > 0, then y = (x0 y 0 )1/2 ,

x = (x0 /y 0 )1/2 ∂x ∂x ∂y ∂y ds2 = dx2 + dy 2 = ( 0 dx0 + 0 dy 0 )2 + ( 0 dx0 + 0 dy 0 )2 ∂x ∂y ∂x ∂y √ √ √ 0 0 0 1 x y x = ( √ 0 0 dx0 − p 3 dy 0 )2 + ( √ dx0 + √ 0 dy 0 )2 0 2 xy 2 y 2 x 2 y0 √ x0 1 = y 0 2 0 2 x0 (1 + y 02 ) 0 2 1 p 0 0 x y − )dx0 dy 0 = dx + dy + ( 2 4x0 y 0 y02 4y 0 3 4

Physics 1. Mechanics

Problem 1.14.

Problems

Calculate the length of the parabola y = ax2 , a = const, from x = −a to x = a.

Solution. 2

2



2

 dy 2 1 + ( ) ) dx2 = (1 + 4a2 x2 )dx2 dx

ds = dx + dy = Z Z a√ s = ds = 1 + 4a2 x2 dx −a

Problem 1.15. Ellipse is given by the equation x2 /a2 + y 2 /b2 = 1. Calculate the circumference. Solution. Let x = a cos t, y = b sin t, 0 ≤ t < 2π. Then   dx 2 dy 2 ds = dx + dy = ( ) + ( ) dt2 = (a2 sin2 t + b2 cos2 t)dt2 dt dt Z Z 2π p s = ds = a2 sin2 t + b2 cos2 tdt 2

2

2

0

ˆ Write down the equation Problem 1.16. A plane passes through the point r0 and its normal is n. of the plane. Solution. All vectors in the plane are perpendicular to the normal, therefore the plane equation is (r − r0 ) · n = 0 Problem 1.17. Find the relation between the unit vectors of 3D Cartesian coordinates and unit vectors of spherical coordinates. Solution. First of all, we have ˆ + sin θ sin ϕyˆ + cos θzˆ rˆ = sin θ cos ϕx ˆ = − sin ϕx ˆ + cos ϕy ϕ ˆ × rˆ = cos θ cos ϕx ˆ + cos θ sin ϕyˆ − sin θzˆ θˆ = ϕ ˆ = a1 rˆ + a2 + θˆ + a3 ϕ, ˆ then Let now x ˆ · rˆ = sin θ cos ϕ a1 = x ˆ · θˆ = cos θ cos ϕ a2 = x ˆ·ϕ ˆ = − sin ϕ a3 = x

5

Physics 1. Mechanics

Problems

ˆ Similarly for yˆ and z. Problem 1.18. Decompose unit vectors of cylindrical coordinates using unit vectors of spherical coordinates. Solution. We know that ˆ = − sin ϕx ˆ + cos ϕyˆ −ϕ

ˆ + sin ϕyˆ ρˆ = cos ϕx

ˆ Like it was in the previous problem, Let, for example, ρˆ = a1 rˆ + a2 θˆ + a3 ϕ. a1 = ρˆ · rˆ = sin θ a2 = ρˆ · θˆ = cos θ ˆ=0 a3 = ρˆ · ϕ ˆ and z. ˆ and similarly for ϕ Problem 1.19. Solution.

Let ω = ω zˆ (cylindrical coordinates). Derive ω × r, ω × (ω × r), and r × (ω × r).

ˆ + y yˆ + z z) ˆ = −ωy x ˆ + ωxyˆ ω × r = ω zˆ × (xx ˆ + y y) ˆ ω × (ω × r) = ω(ω · r) − ω 2 r = ω 2 z zˆ − ω 2 r = −ω 2 (xx ˆ + y y) ˆ r × (ω × r) = ωr2 − r(ω · r) = ω(x2 + y 2 )zˆ − ωz(xx

Problem 1.20. Two cities on Earth surface (assume it is a sphere with the radius R) have the coordinates (α1 , β1 ) and (α2 , β2 ), where α is the latitude, and β is the longitude. Find the shortest distance between the cities. Hint: Arc of the circle passing through the center of the Earth. ~ 1 and OP ~ 2 (O is Solution. The distance l = Rγ, where γ is the angle between the vectors OP the center). One has ~ 1 = R(cos α1 cos β1 , cos α1 sin β1 , sin α1 ) OP ~ 2 = R(cos α2 cos β2 , cos α2 sin β2 , sin α2 ) OP ~ 1 · OP ~ 2 /|OP ~ 1 | · |OP ~ 2 | = cos α1 cos α2 cos(β1 − β2 ) + sin α1 + sin α2 cos γ = OP

Problem 1.21. Find the volume.

Four vertices of the tetrahedron are given by four radius-vectors ri , i = 1, 2, 3, 4.

6

Physics 1. Mechanics

Problems

Solution. V = (1/6)|(r2 − r1 ) · ((r3 − r1 ) × (r4 − r1 ))|

p Problem 1.22. Find the distance element ds2 in the elliptical coordinates r = x2 /a2 + y 2 /b2 , tan ϕ = y/x. Define unit vectors tangential and normal to r = const curves and derive their relation ˆ and y. ˆ to x Solution. As we found earlier, abr cos ϕ

x= p y=p

a2 sin2 ϕ + b2 cos2 ϕ abr sin ϕ

a2 sin2 ϕ + b2 cos2 ϕ

We have further 2



ds =

∂x ∂x dr + dϕ ∂r ∂ϕ

2

 +

∂y ∂y dr + dϕ ∂r ∂ϕ

2

The derivatives are ab cos ϕ ∂x =p 2 ∂r a sin2 ϕ + b2 cos2 ϕ ∂y ab sin ϕ =p ∂r a2 sin2 ϕ + b2 cos2 ϕ   abr sin ϕ ∂x cos2 ϕ(a2 − b2 ) = −p 1+ 2 2 ∂ϕ a sin ϕ + b2 cos2 ϕ a2 sin2 ϕ + b2 cos2 ϕ   ∂y abr cos ϕ cos2 ϕ(a2 − b2 ) =p 1− 2 2 ∂ϕ a sin ϕ + b2 cos2 ϕ a2 sin2 ϕ + b2 sin2 ϕ Please substitute. ˆ Let us consider two close points on the ellipse: Calculation of the tangential unit vector t. (x, y) and (x + dx, y + dy). The vector connecting these two points is a tangential vector, that is, ˆ However, we have to take into account that dx and dy are not independent, ˆ + dy · yˆ k t. dr = dx · x since they have to be calculated with r = const, that is, dx = (

∂x )dϕ, ∂ϕ

dy = (

∂y )dϕ ∂ϕ

Using this we find tˆ =



   ∂x ∂y ∂x 2 ∂y 2 ˆ + ( )yˆ / ( ) + ( ) ( )x ∂ϕ ∂ϕ ∂ϕ ∂ϕ

7

Physics 1. Mechanics

Problems

ˆ ˆ = zˆ × t. Please substitute. After tˆ is found, the normal unit vector can be found as n Problem 1.23. Solution.

Simplify (a × b) · (c × d).

(a × b) · (c × d) = a · (b) × (c × d)) = a · (c(b · d) − d(b · c)) = (a · c)(b · d) − (a · d)(b · c)

Problem 1.24. Given two non-parallel unit vectors e1 and e2 , find another unit vector e3 such that all three are in the same plane and e3 divides the angle between the first two in the way e1ˆe3 = 2e2ˆe3 . Solution. Let eˆ3 = aˆ e1 + bˆ e2 . Since eˆ3 · eˆ3 = 1 we have a2 + b2 + 2abˆ e1 · eˆ2 = 1 Let α be the angle between eˆ1 and eˆ3 , and β be the angle between eˆ2 and eˆ3 . We have cos α = eˆ1 · eˆ3 = a + bˆ e1 · eˆ2 cos β = eˆ2 · eˆ3 = b + aˆ e1 · eˆ2 Since α = 2β we have cos α = cos(2β) = 2 cos2 β − 1 so that eˆ1 · eˆ3 = a + bˆ e1 · eˆ2 = 2(ˆ e2 · eˆ3 = b + aˆ e1 · eˆ2 )2 − 1 Since eˆ1 · eˆ2 is known we get two equations for a and b which have to be solved. Please solve. Problem 1.25.

What is the projection of the vector a onto the unit vector eˆ ?

Problem 1.26.

What is the projection of the vector a onto the vector b ?

Problem 1.27. Given two vectors a and b, represent vector a as a sum of two vectors, ak and a⊥ , a = ak + a⊥ , such that ak k b and a⊥ ⊥ b. Problem 1.28. On the Earth a man is at the point 20◦ east longitude and 40◦ latitude moves in the north-east direction. Express the unit vector along the velocity in spherical and cartesian coordinates. Problem 1.29. Show that a straight line is given by the relation r = λa + b, where a and a are constant vectors and −∞ < λ < ∞. Problem 1.30.

Find the intersection line of the two planes r · a1 = d1 and r · a2 = d2 . 8

Physics 1. Mechanics

Problems

Solution. Two planes intersect by a straight line (we assume that a1 and a2 are not parallel). First let us write the equation of a straight line which passes through the point r0 in the direction given by the vector l. Since for each point r on this straight line the vector (r − r0 ) k l we have (r − r0 ) × l = 0 This is the vector equation for a straight line. Now, our line has to be perpendicular to a1 and a2 , thus we may choose l = a1 × a2 . The point r0 belongs two the both planes simultaneously, that is, r0 · a1 = d1 and r0 · a2 = d2 . Let us try to find is as r0 = k1 a1 + k2 a2 . The conditions give k1 a21 + k2 a1 · a2 = d1 k1 a1 · a2 + k2 a22 = d2 This set of equations always has a unique solution since a21 a22 > (a1 · a2 )2 . Problem 1.31. Write in the vector form the equation for a circle with the radius R, normal ˆ (unit vector !), and center at r0 . direction n Solution. Set of equations: (r − r0 ) · n = 0 (r − r0 )2 = R2

Problem 1.32. Write in the vector form the equation for a cylindrical surface with the radius R and the axis parallel to a and crossing the point r0 . ˆ = a/|a|. Then the equation is Solution. Let a (r − r0 )2 − (ˆ a · (r − r0 ))2 = R2

Problem 1.33. Prove that four different points ri , i = 1, 2, 3, 4, are always on a sphere and find the center and the radius of the sphere. Solution. Let r0 be the center of a sphere and R be its radius. We have (ri − r0 )2 = R2 where i = 1, 2, 3, 4. Subtracting, say, equation for i = 1 from others, we have r0 · X2 = r22 − r12 r0 · X3 = r32 − r12 9

Physics 1. Mechanics

Problems r0 · X4 = r42 − r12

where Xj = rj − r1 , j = 2, 3, 4. We assume that Xj are not all in the same plane. Then we can decompose r0 = k2 X2 + k3 K3 + k4 K4 and substituting one has k2 X2 · K2 + k3 X3 · X2 + k4 X4 · X2 = r22 − r12 k2 X2 · K3 + k3 X3 · X3 + k4 X4 · X3 = r32 − r12 k2 X2 · K4 + k3 X3 · X4 + k4 X4 · X4 = r42 − r12 These equations always have a solution (why ?). Problem 1.34. Problem 1.35. δin δjm . Problem 1.36.

P

Calculate a) Prove a)

i δii ,

P

ijk

b)

P

ij

δij , c)

εijk εijk = 6, b)

P

j

P

jk

δij δjk , d)

P

jk δij δjk δkl .

εijk εmjk = 2δim , c)

P

k

εijk εmnk = δim δjn −

Express A × (B × C) with the use of εikj and δij .

Problem 1.37. Angular momentum is defined as J = mr × v. Express in terms of angular velocity for a circularly moving particle. Problem 1.38. Given two nonparallel vectors a and b build three mutually perpendicular unit vectors. Solution. For example, we may choose eˆ1 = a/|a|, eˆ2 = a × b/|a × b|, and eˆ3 = eˆ1 × eˆ2 .

2

Velocity, acceleration, trajectory

Problem 2.1. A particle moves in the x − y plain according to the law: x = k1 t, y = k2 t2 . Find: a) velocity, b) acceleration, c) distance from the coordinate origin. Problem 2.2. A particle moves along the ellipse x2 /a2 + y 2 /b2 = 1 so that dϕ/dt = const. Find the velocity and acceleration. Find the radius of curvature. Solution. vx = x˙ = k1 ,

vy = y˙ = 2k2 t

ax = v˙ x = 0,

ay = v˙ y = 2k2

r = (x2 + y 2 )1/2 = (k12 t2 + k22 t4 )1/2

10

Physics 1. Mechanics

Problem 2.3. Find v and a.

Problems

Trajectory is given by y = kx2 , k = const, and x = at2 , k = const and a = const.

Problem 2.4. Trajectory is given by r = kϕ and ϕ = ωt, k = const and ω = const. Find v and a. What is the angle between the velocity and acceleration as a function of time ? Solution. x = kωt cos ωt y = kωt sin ωt vx = kω(cos ωt − ωt sin ωt) vy = kω(sin ωt + ωt cos ωt) ax = kω 2 (−2 sin ωt − ωt cos ωt) ay = kω 2 (2 cos ωt − ωt sin ωt) c = a · v/|a| · |v| cos(av) = (ax vx + ay vy )/(a2x + a2y )1/2 (vx2 + vy2 )1/2 Solution. Second method: polar coordinates. r = rrˆ,

ˆ + sin ϕyˆ rˆ = cos ϕx v = r˙ = r˙ rˆ + rrˆ˙ ˆ + cos ϕy) ˆ ϕ˙ = r˙ rˆ + r(− sin ϕx ˆ = r˙ rˆ + rϕ˙ ϕ ˆ˙ = (− cos ϕx ˆ − sin ϕy) ˆ ϕ˙ ϕ = −ϕˆ ˙r ˆ rˆ˙ = ϕ˙ ϕ ˆ˙ ˆ + rϕ¨ϕ ˆ + rϕ˙ ϕ a = v˙ = r¨rˆ + r˙ rˆ˙ + r˙ ϕ˙ ϕ ˆ = (¨ r − ϕ˙ 2 r)ˆ r + (rϕ¨ + 2r˙ ϕ) ˙ ϕ We have ϕ = ωt so that ϕ˙ = ω, ϕ¨ = 0. We have r = kωt, so that r˙ = kω, r¨ = 0. Now ˆ v = kω rˆ + ωrϕ ˆ a = −ω 2 rrˆ + 2kω 2 ϕ

Problem 2.5.

Two spacecraft are orbiting Earth. The orbit radii and angular velocities are the 11

Physics 1. Mechanics

Problems

same but one (A) orbit is always above the equator, while the other (B) passes above the poles. When B is above the equator the spacecraft A is on the opposite side of the diameter. Find the vectors connecting A and B as a function of time. Problem 2.6. Express the velocity vector (in general) in terms of spherical coordinates and unit vectors eR , eθ , eϕ . Solution. ˆ dr = drrˆ + rdθθˆ + r sin θdϕϕ ˆ v = r˙ rˆ + rθ˙θˆ + r sin θϕ˙ ϕ

Problem 2.7. A particle moves according to the law: x = k1 t cos(ωt), y = k2 t sin(ωt), z = bt2 . Find the velocity and acceleration. Solution. vx = k1 cos ωt − k1 ωt sin ωt vy = k2 sin ωt + k2 ωt cos ωt vz = 2bt ax = −2k1 ω sin ωt − k1 ω 2 t cos ωt ay = 2k2 ω cos ωt − k2 ω 2 t sin ωt az = 2b

Problem 2.8. acceleration.

Derive the expression for the trajectory length for the motion with constant

Problem 2.9. A particle moves along the trajectory r = a/(1 −  cos ϕ) in cylindrical coordinates so that r2 (dϕ/dt) = l. Here a, , and l are constant parameters. Find (dr/dt) as a function of r. Same as a function of ϕ. Solution. r˙ =

a sin ϕ l dr ϕ˙ = − 2 dϕ (1 −  cos ϕ) r2

(1 −  cos ϕ) = a/r cos ϕ = (1 − a/r) sin ϕ = ±(1 − (1 − a/r)2 /2 )1/2 r˙ = ∓

l(2 − (1 − a/r)2 )1/2 a 12

Physics 1. Mechanics

Problems

Problem 2.10. Given x(t) = R cos(ωt), y(t) = R sin(ωt). Find the angle between the vectors of velocity and acceleration as a function of time. Solution. vx = −γ1 R exp(−γ1 t) cos(ω1 t) − ω1 R exp(−γ1 t) sin(ω1 t) vy = −γ2 R exp(−γ2 t) sin(ω2 t) + ω2 R exp(−γ2 t) sin(ω2 t) ax = (γ12 − ω12 )R exp(−γ1 t) cos(ω1 t) + 2ω1 γ1 R exp(−γ1 t) sin(ω1 t) ay = (γ22 − ω22 )R exp(−γ1 t) sin(ω1 t) − 2ω2 γ2 R exp(−γ2 t) cos(ω2 t) c = a · v/|a| · |v| cos(av) = (ax vx + ay vy )/(a2x + a2y )1/2 (vx2 + vy2 )1/2

Problem 2.11. Given x(t) = R exp(−γ1 t) cos(ω1 t), y(t) = R exp(−γ2 t) sin(ω2 t). Find the angle between the vectors of velocity and acceleration as a function of time. Problem 2.12. functions of time. Solution.

Given r(t) = kt, ϕ(t) = ωt. Find the tangential and normal acceleration as

x = kt cos ωt,

y = kt sin ωt

vx = k cos ωt − kωt sin ωt vy = k sin ωt + kωt cos ωt ax = −2kω sin ωt − kω 2 t cos ωt ay = 2kω cos ωt − kω 2 t sin ωt vx vy p ˆ vˆ = v/|v| = p 2 x + yˆ vx + vy2 vx2 + vy2 at = (a · vˆ)ˆ v,

an = a − at

Solution. Second method: polar coordinates. As in Problem 0.2 ˆ = k rˆ + rω ϕ ˆ v = r˙ rˆ + rϕ˙ ϕ ˆ a = (¨ r − ϕ˙ 2 r)ˆ r + (rϕ¨ + 2r˙ ϕ) ˙ ϕ ˆ = −ω 2 rrˆ + 2kω ϕ |v| = (k 2 + ω 2 r2 )1/2 at = (a · v)v/|v|2

13

Physics 1. Mechanics

Problems

an = a − at

Problem 2.13. Given vx (t) = vd + v0 cos(ωt), vy (t) = v0 sin(ωt). Find x(t) and y(t). What are the conditions on the parameters for the absence of self-intersection. Problem 2.14. Given vx (t) = vd + v0 cos(ωt), vy (t) = v0 sin(ωt), vz (t) = at. Find the distance from the coordinate origin to the particle as a function of time if r(t = 0) = 0. Solution. Z t v0 vx (t0 )dt0 = vd t + sin ωt x = x0 + ω t0 v0 y = (1 − cos ωt) ω z = at2 /2 r = (x2 + y 2 + z 2 )1/2

Problem 2.15. In a universe all bodies move away from the coordinate origin with the velocities v = Kr, where K = const. What would see an observer at an arbitrary position r0 ? Solution. r 0 = r − r0 v 0 = v − v0 = K(r − r0 ) = Kr 0

Problem 2.16. * A rabbit starts to run at t = 0 from the point (x0 , 0) in the positive direction of axis y with the velocity v0 (magnitude !). A fox starts to run from (0, 0) at the same moment and its velocity v > v0 always points towards the rabbit. How much time does it take to catch the rabbit. Problem 2.17. Solution.

Given ax = a0 exp(−γt), ay = a1 sin(ωt). Find r(t).

ˆ + a1 sin ωtyˆ a = a0 exp(−γt)x Z t v = v0 + a(t0 )dt0 t0

a0 ˆ = v0 − [exp(−γt) − exp(−γt0 )]x γ a1 − (cos ωt − cos ωt0 )yˆ ω Z t

v(t0 )dt0

r = r0 + t0

14

Physics 1. Mechanics

Problems

= r0 + v0 (t − t0 ) +

a0 exp(−γt0 ) ˆ (t − t0 )x γ

a1 cos ωt0 (t − t0 )yˆ ω a0 ˆ + 2 [exp(−γt) − exp(−γt0 )]x γ a1 − 2 [sin ωt − sin ωt0 ]yˆ ω

+

Problem 2.18. Given vx = v0x exp(−γt), vy = v0y sin(ωt), vz = v0z + at. Write down the expression for the path length (integral). Problem 2.19. Given x = x0 cos(ωt), y = y0 sin(ωt), x0 6= y0 , find the tangential and normal components of the acceleration. Problem 2.20. * A body starts from the equator of the sphere (“Earth”) with the radius R and moves all the time in the north-east direction so that the velocity magnitude v remains constant. Where does it stop and how much time does it take ?

3

Inertial and noninertial reference frames

Problem 3.1. A plane takes off with the acceleration 0.5|g| at the angle 30◦ to the horizon. What is the weight of the 75 kg passenger ? Solution. p W = m|g + a| = m (g + a sin θ)2 + a2 cos θ2 Problem 3.2. equator ? Solution.

What should be the length of the day on Earth to compensate the gravity at the

g = ω 2 R,

T =

2π = 2π ω

s

R g

Problem 3.3. A body starts moving with the velocity v from the center of the rotating disk (angular velocity ω). There are no external forces. Describe the motion from the point of view of the rotating observer. Solution. x = vt,

y=0

x0 = x cos ωt − y sin ωt = vt cos ωt 15

Physics 1. Mechanics

Problems y 0 = x sin ωt + y cos ωt = vt sin ωt

Problem 3.4. What is the weight of a standing 1000 kg car on the equator ? What is its weight if it is moving in the east direction with the velocity 300 km/hour ? Solution. W = mg − mω 2 R W = mg − mω 2 R − 2mvω

Problem 3.5. A biker enters a quarter-circle turn of the radius R with the velocity v. What is the angle between the biker’s body and the vertical ? Solution. v2 tan θ = a/g = gr Problem 3.6. A body hangs on a rope from the ceiling in a standing train. The train starts moving with the acceleration a. What is the angle between the rope and the vertical ? Problem 3.7. A body hangs on a rope from the ceiling in a rotating (angular velocity ω) cell. The distance from the rotation center is r. What is the angle between the rope and the vertical ? Problem 3.8. A horizontal carousel rotates with the angular velocity ω. What is the weight of a person who sits at the radius R ? Problem 3.9. A project of a space station suggests rotation in order to produce artificial gravity. If the diameter of the station is 20 m, what should be the rotation period in order to produce the gravity equivalent to 0.5g ? Problem 3.10. A body is moving along x axis with constant velocity vx in the inertial (standing) frame. Write down x0 (t) and y 0 (t) in the rotating frame. What is the direction of acceleration as a function of time in the rotating frame ? Solution. x0 = vx t cos ωt y 0 = vx t sin ωt a0x = x¨0 = −2ωvx sin ωt − ω 2 vx t cos ωt a0y = y¨0 = 2ωvx cos ωt − ω 2 vx t sin ωt

16

Physics 1. Mechanics

Problems

Problem 3.11. A body falls with the velocity v = gτ [1 − exp(−t/τ )] (because of the air drag force). Write down the second Newton law in its frame. Solution. A = v˙ = g exp(−t/τ ) ma = F − mA

Problem 3.12. A river flows from the north to the south in the northern hemisphere at the latitude θ. The flow velocity is v and the river width is L. What is the difference of the water level at the western and eastern coasts ? (Hint: Coriolis.) Solution. ax = 2ω cos θv ay = g h/L = ax /ay =

4

2ωv cos θ g

Particle dynamics, Newton laws

Problem 4.1. A particle is moving so that ϕ = at2 , r = r0 exp(bt), where a, b, and r0 are constants. Find the force. Problem 4.2. A body (mass m) starts falling. The air friction force is Ff = −kv, where k = const and v is the body velocity. Find v(t) and r(t). Solution. v = v zˆ mv˙ = mg − kv kt mg − kv = ln mg m mg v= (1 − exp(−kt/m)) k Problem 4.3. A body (mass m) is thrown horizontally with the initial velocity v0 . The air friction force is Ff = −kv, where k = const and v is the body velocity. Find v(t) and r(t).

17

Physics 1. Mechanics

Problems

Solution. mv˙ x = −kvx mv˙ y = mg − kvy vx = v0 exp(−kt/m) mg (1 − exp(−kt/m)) vy = k Problem 4.4. v(t) and r(t). Solution.

Force Fx = F0 sin2 (ωt) acts on a particle (mass m) which is initially at rest. Find

mv˙ x = F0 sin2 (ωt) = 21 F0 [1 − cos(2ωt)]   1 F0 t− sin(2ωt) x= 2m 2ω

Problem 4.5. At high speeds the air drag force (friction) is Ff = −kvv. A body is falling vertically in the air with the initial velocity v0 . Find v(t) and r(t). Solution. mv˙ = mg − kv 2 Z v kt dv 0 2 = 0 m 0 mg/k − v r 2 |mg/k − v | mg kt ln =2 mg/k k m

Problem 4.6. A charged particle (charge q, mass m) is accelerated by the electric field E = E1 sin(ω1 t)ˆ ex + E2 sin(ω2 t)ˆ ey . Find the trajectory. Problem 4.7. A charged particle moves with constant velocity v ⊥ B (B - magnetic field). Find the electric field. Problem 4.8. A charged particle (mass m, charge q, velocity v ) enters a cylinder with the length l. The entry point is at the cylinder axis, and the particles enters at the angle α to the axis. There is a homogeneous magnetic field along the axis inside the cylinder. At what distance from the axis

18

Physics 1. Mechanics

Problems

the particle leaves the cylinder ? Problem 4.9. A charged particle (mass m, charge q) is at rest in an homogeneous magnetic field B = (0, 0, B). Suddenly, at t = 0 an electric field E = (0, E, 0) is switched on. The electric field is suddenly switched off at t = T /2, where T = 2π/(|q|B/m). Describe the motion of the particle. What is its final energy ?

5

Potential energy, conservation laws

Problem 5.1. For each of the forces given below check whether it is conservative and find the potential energy, if possible: a) Fx = 2yz(1 − 6xyz), Fy = 2xz(1 − 6xyz), Fz = 2xy(1 − 6xyz) b) Fx = y 2 + z 2 + 2(xy + yz + zx), Fy = x2 + z 2 + 2(xy + yz + zx), Fz = y 2 + x2 + 2(xy + yz + zx) Problem 5.2. ?

Potential energy is given by U = a/r2 − b/r. At what r a particle is in equilibrium

Problem 5.3. Potential energy is given by U (x) = k|x| (one-dimensional motion). Find the period of the bound motion of the particle with the energy E > 0. Solution. r 2 (E − k|x|) vx = ± m dx dt = vx Z E/k dx q T =2 2 −E/k (E − k|x|) m Z E/k dx q =4 2 0 (E − kx) m r 8 m√ = E − kx|0E/k k 2 r 8 mE = k 2 Problem 5.4. Force is given by Fρ = a/ρ2 , Fϕ = b sin ϕ/ρ2 (cylindrical coordinates). Is the force conservative ? If yes, find the potential. What is conserved in this force ?

19

Physics 1. Mechanics

Problems

Solution. Let the force be potential, then ∂U ∂ρ 1 ∂U Fϕ = − ρ ∂ϕ ∂Fρ ∂ρFϕ = ∂ϕ ∂ρ b sin ϕ 0 6= − 2 ρ Fρ = −

Not potential. Problem 5.5. Potential energy is given in polar coordinates by U = a cos ϕ/ρ. Find the force. Is angular momentum conserved ? Find the torque at (ρ, ϕ). Solution. 1 ∂U ∂U ˆ ρˆ − ϕ ∂ρ ρ ∂ϕ a sin ϕ a cos ϕ ˆ ρˆ + ϕ = 2 ρ ρ2

F =−

τ =r×F = ρρˆ × ( =

Problem 5.6. Solution.

a sin ϕ a cos ϕ ˆ ρˆ + ϕ) 2 ρ ρ2

a sin ϕ zˆ ρ

A particle orbit is r = a(1 − cos ϕ). Find the central force.

m 2 J2 r˙ − 2 2mr2 r˙ = a sin ϕϕ˙

U (r) = E −

r˙ 2 = a2 sin2 ϕϕ˙ 2 2

2

= [a − (a − r) ]



J mr2

2

Problem 5.7. A bead (mass m) is moving on a circularly shaped wire (r = const) without friction and is connected to the two points, P1 = (0, −r/2) and P2 = (0, r/2), with identical springs (spring constant k) of initially zero length (so that |F | = kl where l is the length of the spring). a) Write down the force vectors. b) Derive the potential energy (if the forces are conservative). c) Find the

20

Physics 1. Mechanics

Problems

velocity as a function of angle ϕ (for a given energy). c) Find the angular momentum relative to the coordinate origin as a function of ϕ. Solution. Let r1 = (0, −r/2), r2 = (0, r/2) = −r1 F1 = −k(r − r1 ) F2 = −k(r − r2 )  k (r − r1 )2 + (r − r2 )2 U= 2  k 2 = 2r + 2r12 2 Problem 5.8. A bead (mass m) is moving on an elliptically shaped (r = p/(1 −  cos ϕ) wire without friction. The bead is attracted to the focus (0, 0) by the force inversely proportional to the distance r1 squared between the bead and the focus, |F |1 = k1 /r12 . The bead is attracted to the center of the ellipse by the force proportional to the distance r2 between the bead and the center, |F |2 = k2 r2 . a) Write down the force vectors. b) Derive the potential energy (if the forces are conservative). c) Find the velocity as a function of angle ϕ (for a given energy). c) Find the angular momentum relative to the coordinate origin as a function of ϕ. Problem 5.9. In a galaxy the gravitational potential (potential energy) is U = −k/rα , 0 < α < 1. Find the relation between the total energy and angular momentum for circular orbits. Find the dependence of the orbit period on the radius. Solution. J2 k Uef f = − α + r 2mr2 dUef f r˙ = 0 ⇒ =0 dr E = Uef f

Problem 5.10. A particle moves under the influence of the body O which is in the coordinate origin. In the beginning the particle is at a very large distance from O, moves with the velocity v and would pass at the distance l from O if there were no interaction (this is called impact parameter ). What is the minimal distance between the particle and O for the potential energy is U = k/rα (analyze k > 0 and k < 0). Solution. J = mvl

21

Physics 1. Mechanics

Problems

E=

mv 2 J2 =U+ 2 2mr2

Problem 5.11. Find x(t) for a particle with E = 0 in the potential energy U (x) = −ax2 /2+bx4 /4. (Hint: x → 0 for t → −∞.) Solution. p x˙ = ± 2(E − U )/m p = ± 2(ax2 /2 − bx4 /4)/m Z dx p t= 2(ax2 /2 − bx4 /4)/m

Problem 5.12. A satellite of the mass m, moving in the Earth potential U (r) = −k/r, has the total energy E and angular momentum J. Find the maximum (apogee) and minimum (perigee) distance from the Earth. Problem 5.13. A particle (mass m) is moving in the central field U = −k/r on a circular orbit r = r0 . The energy and angular momentum suddenly are changed by ∆E and ∆J. What are the maximal and minimal distances from the attracting body on the new orbit ?

6

Oscillations

Problem 6.1. Find the frequency of small oscillations of a particle (mass m) near the equilibrium in the potential U (x) = a/x13 − b/x7 . Solution. Equilibrium: dU/dx = 0 → −13a/x14 + 7b/x8 = 0 → x0 = (13a/7b)1/6 . Taylor 9 2 15 9 expansion: d2 U/dx2 = 182a/x15 0 − 56b/x0 . Now ω = (182a/x0 − 56b/x0 )/m. Problem 6.2. A particle is in the stable equilibrium in the potential energy U (x) = U0 [1 − 2 2 2 l /(l + x )]. Suddenly it gets a small addition of energy E 0 . Assuming that the oscillations are small find the frequency and amplitude. Problem 6.3. A particle moves in a well of the shape y = ax2 without friction (potential energy U = mgy). Show that the motion can be described as a harmonic oscillation and find the frequency. Solution. Potential energy U = mgy = mgax2 . Kinetic energy K = (m/2)(x˙ 2 + y˙ 2 ) =

m (1 + 4a2 x2 )x˙ 2 2

22

Physics 1. Mechanics

Problems

When ax  1 we have K ≈

m 2 x˙ 2

and ω 2 = 2ga.

Problem 6.4. A body with the mass m is attached to a spring (spring constant k). The other end of the spring is brought into the motion according to the law x = x0 cos(ωt). The friction acting on the body is Fx = −bvx . Show that the body can oscillate with a constant amplitude and find this amplitude. Solution. Let y be the coordinate of the body. Then F = −k∆l = −k(y − x). We have m¨ y = −k(y − x) − by˙ ⇒ m¨ y + by˙ + ky = kx0 cos(ωt) Therefore, y = A cos(ωt − φ). Substitute and complete. Problem 6.5. Find the motion of an oscillator with the natural frequency ω0 and mass m under the force F = F0 + F1 cos(ωt), ω 6= ω0 . Solution. If F = F0 the solution is x1 = F0 /mω 2 . When F = F1 cos(ωt) the solution is x2 = A cos(ωt − φ). The total solutions is x1 + x2 . Problem 6.6. Find the frequency of small radial oscillations of a particle with a mass m near a circular orbit r = r0 in a central potential U (r) = −k/r. Solution. Uef f = −k/r + J 2 /2mr2 . Equilibrium: k/r2 − J 2 /mr3 = 0 ⇒ r0 = J 2 /km. Taylor: d2 Uef f /dr2 = −2k/r03 + 3J 2 /mr04 and ω 2 = (−2k/r03 + 3J 2 /mr04 )/m. Problem 6.7. Find the frequency of small radial oscillations of a particle with a mass m near a circular orbit r = r0 in a central potential U (r). Problem 6.8. Find the average power of the external force F = F0 cos(ωt) for the oscillator: 2 m¨ x + Γx˙ + mω0 x = F (t). Solution. Substitute x = C1 cos(ωt) + C2 sin(ωt): (ω02 − ω 2 )(C1 cos(ωt) + C2 sin(ωt)) + ωΓ(−C1 sin(ωt) + C2 cos(ωt)) = F0 cos(ωt)/m (ω02 − ω 2 )C1 + ωΓC2 = F0 /m (ω02 − ω 2 )C2 − ωΓC1 = 0 C1 = (ω02 − ω 2 )C2 /ωΓ F0 ωΓ C2 = 2 m[(ω0 − ω 2 )2 + Γ2 ω 2 ] P = F vx = F0 cos(ωt)ω(−C1 sin(ωt) + C2 cos(ωt)) hP i = 21 F0 C2 ω hP i =

F02 ω 2 Γ 2m[(ω02 − ω 2 )2 + Γ2 ω 2 ] 23

Physics 1. Mechanics

Problems

Problem 6.9. A bead of the mass m can move on a straight wire along y axis without friction. The bead is connected to two springs (spring constant k, length a). The springs are connected to the points (−l, 0) and (l, 0), l > a, respectively. Initially the bead starts moving from (0, 0) with the velocity v. Assuming that the the oscillations are small, find their frequency and amplitude. Solution. p my˙ 2 + 2 · 12 k( l2 + y 2 − a)2 2 my˙ 2 1 y 2 (l − a) ≈ + 2 · 2k[(l − a)2 + ] 2 l 2k(l − a) ω2 = ml E=

Problem 6.10. A particle is moving in the magnetic field B = (0, 0, B), B = const, and electric field E = (E cos(ωt), 0, 0), ω 6= |q|B/m. Find v(t). mv˙ x = qE cos(ωt) + qvy B mv˙ y = −qvx B ⇒ vx = −(m/qB)v˙ y −(m2 /qB)v¨y = qE cos(ωt) + qvy B v¨y + Ω2 vy = F cos(ωt),

Ω = qB/m,

F = −Ω2 E/B

Substitute vy = A cos(ωt − φ). Problem 6.11. A particle with the mass m is moving in x − y plane with the potential energy U = 21 k(x2 + y 2 ). Initially the particle is in the position (a, 0) and its velocity is (0, v0 ). Find the trajectory.

7

Rotation of a rigid body

Problem 7.1. A disk of the mass m and radius r is connected to two parallel identical springs (k, l) as shown in the figure.

24

7

Rotation of a rigid body

Physics 1. Mechanics Problem 7.1. A disk of the mass m and radius r

is connected to two parallel identical springs

Problems

(k, l) as shown in the figure.

d

l

d

l

Find the frequency of rotational oscillations around the center of the disk.

Find the frequency of rotational oscillations around the center of the disk. Solution. Let the disk rotate by a small angle θ  1. The total energy is E = I θ˙2 /2+2· 21 k(dθ)2 , 22 so that ω 2 = 2kd2 /I, where I = mr2 /2. Problem 7.2. A ball of the radius r rotates with the angular velocity ω around the horizontal axis passing through the center-of-mass. The ball is carefully put on a horizontal surface with the friction coefficient µ. Find vcm (t). Solution. The ball starts to move with sliding so that the friction force F = µmg accelerates the ball but the torque N = F r = µmgr decelerates the rotation: mv˙ = µmg ⇒ v = µgt I ω˙ = −µmgr ⇒ ω = ω0 − (µmgr/I)t where I = (2/5)mr2 . When v = ωr the sliding stops, the friction force vanishes and the ball continues to roll without sliding so that v and ω no longer change. This happens when µgt = ω0 r − (µmgr2 /I)t ⇒ t = ω0 r/µg(1 + mr2 /I) and the velocity is v = ω0 r/(1 + mr2 /I) Problem 7.3. A cylinder of the radius a is rolling without sliding inside a larger cylinder of the radius R as shown in the figure (vertical cross-section).

25

Problem A cylinder of the radius a is rolling without friction inside a larger cylinder of the Physics 1.7.3. Mechanics Problems radius R as shown in the figure (vertical cross-section).

a) Find the minimal angular frequency in the lowest point which allows to reach the highest point. a)b)Find thethe minimal angular frequency in the near lowest which allows to reach the highest point. Find frequency of small oscillations thepoint equilibrium. b)Solution. Find the frequency small oscillations thecylinder equilibrium. When the of center-of-mass of thenear small deviates by θ from the vertical position,

When the − center-of-mass of the smallitscylinder deviates theis vertical position, it Solution. rotates by ϕ = Rθ/a θ = (R − a)θ/a around center of mass. by Theθ from energy therefore given as follows I (R − a)2 θ˙2 m + mg(R − a)(1 − cos θ) E = (R − a)2 θ˙2 + 23 2 2 a2 Complete the solution. Problem 7.4. A homogeneous cube is rotating around the axis passing through the center-ofmass. Describe qualitatively the motion of the axis depending on the angle of the axis with the normal to the cube side. Solution. The axis does not move. Problem 7.5. A hollow cylinder and a solid cylinder of the same radius start rolling simultaneously without sliding down the same slope from the same height. What is the ratio of the final velocities ? Which one comes to the end of the slope earlier and what is the ratio of times ? Solution. For the rolling without sliding one has mgr sin α = (Icm + mr2 )a/r that is, a=

g sin α 1 + Icm /mr2

Since v 2 = 2al, where l is the length of the slope, one has v ∝

√

√ a. Respectively, t ∝ 1/ a.

Problem 7.6. Two identical masses m connected by a massless rod of the length l are moving on a circular orbit r = const around the Earth. The attraction force between the Earth and a point mass 26

Physics 1. Mechanics

Problems

is |F | = GM m/r2 , where M is the Earth mass and G is a universal constant. Find the frequency of small rotational oscillations of the system (masses on the rod) around the center-of-mass. Solution. The equilibrium position is along the radius, when the center of mass moves on the circular orbit. The angular velocity ω0 on the orbit is obtained from mω02 r = GM m/r2 , that is, p ω0 = GM/r3 . Since the graviational force and the centrifugal force depend on the radius, different forces are applied to the two masses. The total force on a mass at the radius R, rotating with the rod, is F = −GM m/R2 + mω 2 R, so that the difference is ∆F = (dF/dR)l = (2GM m/r3 + ω02 )l = (3GM m/r3 )l. If the rod rotates by a small angle θ the resulting torque is N = −∆F lθ, so that one has I θ¨ = −(3GM ml2 /r3 )θ where I = 2m(l/2)2 = ml2 /2. Thus, Physics 1. Mechanics

ω 2 = 6GM/r3 = 6ω02

Problems

Problem 7.7. A car engine is applying a torque to a wheel. The wheel mass is m, radius is r and Problem 7.7. A car engine is applying a torque to a wheel. The wheel mass is m, radius is r and the moment of inertia with respect to the center is I. The coefficient of the static friction with the the moment of inertia with respect to the center is I. The coefficient of the static friction with the road is µ. What is the maximum torque N which can be applied without making the wheel slide ? road is µ. What is the maximum torque N which can be applied without making the wheel slide ? Solution. Letthe thefriction friction force f , fthen = ma r = Iα = Ia/r.a one Excluding a one Solution. Let force be f ,be then = maf and N −and f r =NIα−=f Ia/r. Excluding 2 2 2 ) ≤ µmg ⇒ N ≤ µg(1 + gets rm/(I++mr mr getsff= =N Nrm/(I ) ≤ µmg ⇒ N ≤ µg(1 + I/mr 2 ).I/mr ). Problem 7.8. 7.8. AAbobbin is moved by pulling a thread which is which winded is onwinded the inneron cylinder. The cylinder. The Problem bobbin is moved by pulling a thread the inner outer radius is R, the inner radius is r, the bobbin mass is m, the friction coefficient is µ. What is outer radius is R, the inner radius is r, the bobbin mass is m, the friction coefficient is µ. What is the maximal force F for which the bobbin rolls without friction ? What is the bobbin velocity after the maximal force F for which the bobbin rolls without friction ? What is the bobbin velocity after it moves by the distance l from the rest ? it moves by the distance l from the rest ?

Solution. See previous problem. Here N = F r. Solution. See previous problem. Here N = F r.

Problem Sixidentical identical point masses are positions in the positions 0, 0), r2 = (−a, 0, 0), 1 = Problem 7.9. 7.9. Six point masses m are m in the r1 = (a, 0,r0), r2 (a, = (−a, 0, 0), r3r3==(a, 0), rr4 4==(−a, (−a, (0, r0,6 a), r60,= (0, The 0, −a). Thevelocity anglular velocity (a,a, a, 0), −a,−a, 0), 0), r5 =r5(0,=0, a), = (0, −a). anglular vector ω = vector ω = (ω, 0, 0). Find J. Problem 7.10.

Three identical disks with the mass m and radius r each are connected so that

they have the common center and their planes are mutually perpendicular. Find the moment of inertia relative to an arbitrary axis passing through the center. Solution. Let us choose axes x, y, z along the disks axes. Then for one of them Ixx ≡ I0 = 12 mr 2 ,

27

Physics 1. Mechanics

Typical problems for exam

Problems

(ω, 0, 0). Find J . Problem 0.1. moving so disks that ϕ with = at2 ,the r =mass r0 exp(bt), where a, b, and r0 are Problem 7.10. A particle Threeis identical m and radius r each are connected so that constants. force. they haveFind thethe common center and their planes are mutually perpendicular. Find the moment of inertia relative to an arbitrary passing through the Ucenter. Problem 0.2. A particle (mass m)axis is moving in the central field = −k/r on a circular orbit r =Solution. r0 . The energy changed ∆E and ∆J.for What Letand usangular choosemomentum axes x, y,suddenly z alongarethe disksbyaxes. Then oneare of the them Ixx ≡ I0 = 21 mr2 , and minimal distances from the attracting body on the new orbit ? Imaximal yy = Izz = I0 /2 (other components). For the others one just has to change x to y or to z. Thus, the total tensor of inertiaisisindiagonal 2I0 = energy mr2 , that is,UI0ij[1 − = mr2 δij . This means xx = Iyy = zz = Problem 0.3. A particle the stable Iequilibrium in Ithe potential U(x) = l2 /(l2the + x2 )]. Suddenlyofitinertia gets a small addition of energy E ! . Assuming the oscillations small that moment with respect to arbitrary axisthat passing throughare the common center is mr2 . find the frequency and amplitude.

Problem 7.11. Two identical particles of the mass M are connected to the two ends of a Problem 0.4. Two identical particles of the mass M are connected to the two ends of a rigid rigid massless of a. theThe length Therotates system initially rotates around center-of-mass with the massless rod of therod length systema. initially around the center-of-mass with thethe angular velocity ω.velocity One of the a third encounters one (with the a same mass) at rest, which angular ω.particles One ofencounters the particles third one (with themomensame mass) at rest, which tarily sticks to it. What is the angular velocity of the rotation around the center-of-mass after the momentarily sticks to it. What is the angular velocity of the rotation around the center-of-mass after collision ? (No gravity.) the collision ? (No gravity.)

Problem 7.12.A homogeneous A homogeneous (massr) m, radius is struckforce by Fa inhorisontal force F in the Problem 0.5. ball (massball m, radius is struck by r) a horisontal the point which the center by the distance < r. The time force action veryforce small,action but point whichis above is above the center by the l distance l < tr.of The time ist of is very small, but F t is nonzero. Find the velocity of the center-of-mass and the angular velocity of the ball around F t is nonzero. Find the velocity of the center-of-mass and the angular velocity of the ball around the axis which goes through the center-of-mass, if a) there is no friction with the floor, and b) the axis which goes through the center-of-mass, if a) there is no friction with the floor, and b) if the friction prevents sliding. (I2 CM2 =1 25 mr 2 ) friction prevents sliding. (ICM = 5 mr )

Problem 0.6. A source of light is moving with the angular velocity ω on the circle wit 28 radius R in the plane x − y. The light frequency in the source frame is f . A distant observer (

same x − y plane) measures the frequency of received light as a function of time. Find this fun

Physics 1. Mechanics

8

Problems

Gravitation

Problem 8.1. What force acts on a star inside a spherically symmetric galaxy of the mass M and radius R. The star has a mass m and is at bin the radius r < R from the center of the galaxy. Solution. As if all stars within r were in the center: F = GM (r3 /R3 )m/r2 . Problem 8.2. A binary stellar system consists of two identical stars rotating around the centerof-mass of the system on circular orbits. The period of rotation T and the velocity of the stars v are known. Find the masses and the distance between the stars. Solution. l = v/2πT F = Gm2 /l2 = mv 2 /(l/2)

Problem 8.3. Saturn rings consist of football ball size particles which are moving on circular orbits around the planet. What is the maximal ratio of the ring width to its inner radius if the velocities at the inner and outer edge should not differ by more than 0.5% ? Solution. mv 2 /r = GM m/r2 → r = GM/v 2 → ∆r/r = 2∆v/v. Problem 8.4. Three identical stars with the mass m are rotating so that they form an equilateral triangle (side length a). What is the angular velocity ? What is the ratio J/E ? Is this configuration stable ? Solution. F = 2(Gm2 /a2 ) cos 30◦ F = mω 2 (2a cos 30◦ /3) p ω = 3Gm/r3 J = Iw,

I = 3m(2a cos 30◦ /3)2

K = Iw2 /2 U = −3Gm2 /a

Problem 8.5. A particle is moving along the axis of a homogeneous ring (mass M , radius R). The particle velocity at infinity is zero. What is its velocity when it passes through the center of the ring ?

29

Physics 1. Mechanics

Problems

Solution. √ U (z) = −GM m/ R2 + z 2 E = 0 ⇒ z = 0 → mv 2 /2 = GM m/R

Problem 8.6. The space between the two concentric spheres with the radii a and b, a < b, is filled with a matter with the constant density ρ. Find the gravitational field g as a function of radius r in the whole space. Solution. g = F/m r b ⇒ M = 43 πρ(b3 − a3 ) g = GM/r2

Problem 8.7. A binary system consists of two stars with the masses M and 2M . The distance between them is R. Find the period of the orbital motion. Solution. F = 2GM 2 /R2 µ = m1 m2 /(m1 + m2 ) = 2M/3 F = µω 2 R

9

Relativity

Problem 9.1. A box has the volume V in its rest frame. What is its volume in the frame moving with the velocity v0 ? Solution. One dimension is contracted by γ, so that V 0 = V /γ. Problem 9.2. Two bodies are moving with the same speed v. The angle between their velocities is α. What is the relative velocity ? Solution. v0 = (−v, 0, 0)

30

Physics 1. Mechanics

Problems v 0 = (v cos α, v sin α, 0) p γ = 1/ 1 − v 2 /c2 vx0 + v0x v(cos α − 1) = 0 2 1 + vx v0x /c 1 − v 2 cos α/c2 vy0 v sin α = vy = 0 2 γ(1 + vx v0x /c ) γ(1 − v 2 cos α/c2 )

vx =

Problem 9.3. The momentum of a particle is p and the energy is E. Find the velocity v. Solution. v = pc2 /E Problem 9.4. A star is moving on a circular orbit with the radius R and period T . The star emits at the wavelength λ in its rest frame. The line of sight of a distant observer (distance  R) is in the orbit plane (makes the angle 90◦ with the normal to the orbit plane). Find λ0max − λ0min /λ (λ0 is the wavelength measured by the distant observer). Solution. v = 2πR/T s 1 + v/c λmax = λ 1 − v/c s λmin 1 − v/c = λ 1 + v/c

Problem 9.5. Same as in the previous problem but the angle between the line of sight and the orbit plane is θ < 90◦ . Solution. v = 2πR/T λmax = γ(1 + v cos θ/c) λ λmin = γ(1 − v cos θ/c) λ p γ = 1/ 1 − v 2 /c2

Problem 9.6. A hot gas of identical atoms emitting at the same wavelength λ0 (in the rest frame) is observed by a distant observer. The atoms are moving and the number of atoms moving with the velocity v (parallel to the line of sight) is given but the relation N (v) = C exp(−v 2 /2vT2 ), where C and vT are constant. Each atom emits the energy E. How much energy obtaines the distant observer 31

Physics 1. Mechanics

Problems

at the wavelength λ. Problem 9.7. In the rest frame of the particle A, the particle B moves with the velocity vB , and the particle C moves with the velocity vC . What is the angle between the velocities of A and C in the rest frame of B ? Problem 9.8. In the process of β-decay a neutron n decays in into a proton p, electron e− and antineutrino ν¯: n → p + e− + ν¯. The rest masses obey the following inequality mn > mp + me , the rest of the antinuetrino mν = 0. What is the energy of the antineutrino if the proton momentum is negligible ? Solution. pν = −pe → |pν | = |pe | = p p mn c2 = mp c2 + me c4 + p2 c2 + pc

Problem 9.9. What should be the electron Lorentz factor γ in order that the gyroradius of the gyration around the magnetic field B = 10−3 T be equal to the Earth radius R = 6400 km? Find yourself all relevant data. Solution. p˙ = qv × B p = mvγ,

γ = const

mγ v˙ = qv × B ⇒ ω = qB/mγ r = v/ω ≈ c/ω

Problem 9.10. A charged particle of the charge q and mass m is moving in the electric field E = E0 sin(ωt). Find γ(t) if the particle is in the rest at t = 0. Solution. p˙ = qE0 sin(ωt) qE0 (1 − cos(ωt)) p= ω p γ = p2 /m2 c2 + 1

Problem 9.11.

A source of light is moving with the angular velocity ω on the circle with the

32

Physics 1. Mechanics

Problems

radius R in the plane x − y. The light frequency in the source frame is f . A distant observer (in the same x − y plane) measures the frequency of received light as a function of time. Find this function.

10

Problems given at various tests, including midterms and finals

Problem 10.1. A particle of the mass m is moving on the elliptic orbit r = p/(1 −  cos ϕ). The particle is attracted to the focus O1 by the force |F1 | = k1 /r12 (where |r1 | is the distance between the focus O1 and the particle). In addition, it is attracted to the ellipse center O2 with the force |F2 | = k2 r2 . The third force acting on the particle is always perpendicular to the particle velocity. When the particle is in A, its velocity is v1 = (0, v1 ). What is its angular momentum in B ? y B

A

O1

O2 x

Solution. Points: First we need the coordinates of the points A, B, O1 , and O2 : 1. O1 is the coordinate origin, so that rO1 = xO1 = yO1 = 0. 2. A corresponds to ϕA = π, so that rA = p/(1 + ), xA = −p/(1 + ), yA = 0. 3. It is convenient to use the point C: ϕC = 0, rC = p/(1 − ), xC = p/(1 − ), yC = 0. 4. O2 is in the middle between A and C, so that xO2 = (xA + xC )/2 = p/(1 − 2 ), yO2 = (yA + yC )/2 = 0. 5. B is just above O2 , so that xB = xO2 = p/(1 − 2 ). The equation r = p(1 −  cos ϕ) can be written as r − r cos ϕ = p → r = p + x, x = r cos ϕ so that rB = p/(1 − 2 ). Now yB =

q

2 rB − x2B = √

p 1 − 2

Force vectors: We denote the position of the particle with r = (x, y), respectively, other points rO1 = (0, 0), rO2 = (xO2 , 0), rA = (xA , 0), rB = (xB , yB ).

33

Physics 1. Mechanics

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Now we introduce the vectors r1 = r − rO1 = r and r2 = r − rO2 . With this notation the forces (both attractive) will be written as follows F1 = −

k1 rb1 , r12

F2 = −k2 r2 rb2 where rb1 and rb2 are unit vectors, r1 = |r1 | and r2 = |r2 |. Potential energy: Both forces are central forces: the direction of F1 is to O1 and the magnitude depends only on |mbr1 |, the direction of F2 is to O2 and the magnitude depends only on |mbr2 |. Central forces are conservative, and the relation to the potential energy is the following: F1 = −

∂U1 rb1 , ∂r1

F2 = −

∂U2 rb2 ∂r2

which gives ∂U1 k1 = 2, ∂r1 r1

∂U2 = k 2 r2 ∂r2

and finally U1 = −

k1 , r1

U2 =

k2 r22 2

Energy conservation: Since the third (unknown) force is always perpendicular to the velocity, F3 ⊥ v, it does not produce work, so that the energy conservation gives mv 2 mv 2 k1 k2 r22 + U1 + U2 = − + = E = const 2 2 r1 2 When the particle is in the point A, we have v = (0, v1 ), r1 = p/(1 + ), r2 = p/(1 − 2 ). When the particle is in the point B, we have v = (0, v) (this v is the unknown we are looking √ for), r1 = p/(1 − 2 ), r2 = p/ 1 − 2 . Therefore, we have mv12 k1 (1 + ) k2 p 2 mv 2 k1 (1 − 2 ) k2 p 2 − + = − + 2 p 2(1 − 2 )2 2 p 2(1 − 2 ) Finally  1/2 2k1 (1 − ) k2 2 p 2 2 v = v1 − + mp (1 − 2 )2 Angular momentum:

34

Physics 1. Mechanics

Problems

J = mr × v. In the point B the velocity is v = (v, 0, 0), so that mvp Jz = −mvyB = − √ 1 − 2 Problem 10.2. A particle, initially resting in the coordinate origin, suddenly breaks up into three particles with the masses m1 , m2 , and m3 . The particle m1 has the charge q > 0. It starts moving into negative x-direction in the homogeneous magnetic field B = (0, 0, B). After having completed half a circle the particle finds itself at the distance l1 from the starting point. The particle m2 has the charge −q < 0. It starts moving in the positive x-direction, and after having completed half a circle is at the distance l2 from the starting point. What is the velocity of the third particle ? The magnetic force is F = qv × B. y

m1

m2

x

Solution. For each particle in the magnetic field ma = qv × B. Since v ⊥ B, we can write m|a| = |q||v||B|, or mv 2 = |q|B r which gives |q|Br v= m From this expression we immediately find (l = 2r for a semicircle !) v1 =

qBl1 , 2m1

v2 =

qBl2 2m2

Momentum conservation gives m1 v1 + m2 v2 + m3 v3 = 0

35

Physics 1. Mechanics

Problems

or v3 = −

m1 v1 + m2 v2 m3

Substituting v1 = v1 eˆx , v2 = −v2 eˆx , we find v3 =

qB(l2 − l1 ) m2 v2 − m1 v1 eˆx = eˆx m3 2m3

Problem 10.3. A particle of the mass m is connected with the point O with the use of a massless rod of the length L. The rod can freely rotate around O. When the angle between the rod and the ˆ a) What is conserved ? b) Find vθ and vϕ (or vertical is θ0 the particle velocity is v0 = v1 θˆ + v2 ϕ. θ˙ and ϕ) ˙ as functions of θ; c) Find θ¨ and ϕ¨ as functions of θ; d) Find the maximum speed |v|max of the particle; e) Find the tension in the rod as a function of θ.

g↓

θ O

Solution. Preparations: F = −mg zˆ + T rˆ ˆ v = vθ θˆ + vϕ ϕ ˆ = rθ˙θˆ + r sin θϕ˙ ϕ r˙ = 0 a): mv 2 + U, U = mgz = mgr cos θ 2 m E = (vθ2 + vϕ2 ) + mgr cos θ 2 m 2 ˙2 = (r θ + r2 sin2 θϕ˙ 2 ) + mgr cos θ 2 E=

36

Physics 1. Mechanics

Problems ˆ =0 E˙ = T · v = T rˆ · (vθ θˆ + vϕ ϕ) m E = const = (v12 + v22 ) + mgL cos θ0 2 p˙ = F 6= 0 ˆ J = r × mv = mrrˆ × (vθ θˆ + vϕ ϕ) ˆ − mrvϕ θˆ = mrvθ ϕ J˙ = r × F = −rmg rˆ × zˆ J˙z = 0 Jz = J · zˆ = mr sin θvϕ = mr2 sin2 θϕ˙ Jz = const = mv2 L sin θ0

b): Jz mr sin θ Jz vϕ = ϕ˙ = 2 r sin θ mr sin2 θ m 2 (v + v 2 ) + mgr cos θ = E 2 θr ϕ 2E vθ = ± − vϕ2 − 2gr cos θ m r 2E J2 =± − 2 2 z 2 − 2gr cos θ m rm r sin θ vθ 1 2E J2 θ˙ = =± − 2 2 z 2 − 2gr cos θ r r m m r sin θ

vϕ =

c): dθ˙ ˙ θ¨ = ·θ dθ ! r d 1 2E Jz2 = ± − 2 2 2 − 2gr cos θ dθ r m m r sin θ ! r 1 2E Jz2 · ± − 2 2 2 − 2gr cos θ r m m r sin θ  −1/2 1 2E Jz2 − 2 2 2 − 2gr cos θ =± r m m r sin θ  2  Jz cos θ · + gr sin θ θ˙ m2 r2 sin3 θ  2  1 Jz cos θ = 2 + gr sin θ r m2 r2 sin3 θ

37

Physics 1. Mechanics

Problems

dϕ˙ ˙ ·θ dθ r   Jz2 2Jz cos θ 1 2E − 2 2 2 − 2gr cos θ =∓ r m m r sin θ mr2 sin3 θ

ϕ¨ =

d): v2 =

2 (E − mgr cos θ) m

d 2 v = 2gr sin θ > 0 dθ v = vmax → θ = θmax > π/2 θ = θmax ⇒ θ˙ = 0 ⇒ 2E J2 − 2 2 z2 − 2gr cos θmax = 0 m m r sin θmax

e): mv 2 r mv 2 2E − 2mgr cos θ T − mg cos θ = − =− r r 2E T = 3mg cos θ − r (T + mg) · rˆ = −

Problem 10.4. A particle moves according to r = r0 exp(ϕ) in the central force. Find U (r). Solution. In the central force energy and angular momentum are conserved. Energy conservation gives E = 12 mr˙ 2 + 21 mr2 ϕ˙ 2 + U = const. Angular momentum conservation gives: mr2 ϕ˙ = J = const. and therefore ϕ˙ =

J . mr2

From the given relation r = r0 exp(ϕ) we have r˙ = (d/dt)(r0 exp(ϕ)) = r0 exp(ϕ)ϕ˙ = rϕ˙ =

J . mr 38

Physics 1. Mechanics

Problems

Substituting (??) and (??) into the energy conservation (??) we finally obtain E=

J2 +U mr2

and U =E−

Problem 10.5.

J2 mr2

A particle moves according to x = A exp(−γt) cos(ωt) y = A exp(−γt) sin(ωt)

Find tangential acceleration. Solution. The problem is solved by direct differentiating: vx = x, ˙ vy = y, ˙ ax = v˙ x , ay = v˙y , so that the velocity and acceleration are: vx = x˙ = −γA exp(−γt) cos(ωt) − ωA exp(−γt) sin(ωt) = −γx − ωy, vy = −γy + ωx, ax = −γvx − ωvy = (γ 2 − ω 2 )x + 2γωy, ay = −γvy + ωvx = (γ 2 − ω 2 )y − 2γωx. Tangential acceleration is parallel to the velocity vector, so that ak = a · v/|v|: ak = (ax vx + ay vy )/(vx2 + vy2 )1/2 p = −γ γ 2 + ω 2 A exp(−γt)

Problem 10.6. A particle of the mass m moves in the potential energy U (x) = −Ax2 /2 + Bx3 /3, A > 0, B > 0. Find the frequency of small oscillations. Solution. Equilibrium is where Fx = −(dU/dx) = 0: −

dU = −Ax + Bx2 = 0 → x1 = 0, x2 = A/B. dx

Whether the equilibrium is stable or not is determined by the second derivative (d2 U/dx2 ) = −A + 2Bx. If (d2 U/dx2 ) < 0 the point is a maximum and the equilibrium is unstable, if (d2 U/dx2 ) > 0 the point is a minimum and the equilibrium is stable. In x = x1 = 0 we have (d2 U/dx2 ) = −A < 0 so that this is an unstable equilibrium. In x = x2 = A/B we have (d2 U/dx2 ) = A > 0 and this is a 39

Physics 1. Mechanics

Problems

stable equilibrium. Near the equilibrium point the potential energy can be Tayor expanded: U (x) ≈ U (x2 ) + 21 A(x − x2 )2 . Let us denote X = x − x2 then vx = x˙ = X˙ and the energy conservation is written as 1 mX˙ 2 2

+ 21 AX 2 = const

which immediately gives (according to the general rule) ω 2 = A/m.

Problem 10.7. A hollow cylinder is sliding without friction (no rolling) with the velocity v. The cylinder comes to a surface with friction. What is the final velocity of the cylinder ? Solution. When the cylinder comes to the surface with friction it is decelerated by the friction force and at the same time its rotation is acceletrated until the cylinder begins to roll without sliding. In the rolling state the friction force is zero and the velocity does not change. Let the friction force magnitude be Fs . Then the deceleration is mV˙ = −Fs → V (t) = v −

Fs t, m

whereV (t) is the velocity of the center-of-mass in the moment t. The moment of the friction force (torque) accelerates the rotation around the axis passing through the center-of-mass, as follows: Fs r t, I

I ω˙ = Fs r → ω(t) =

where ω(t) is the angular velocity of rotation and I is the moment of inertia around the axis passing P through the center-of-mass. According to the definition I = mi ri2 , where ri is the distance from the rotating mass (small part of the body) from the rotation axis. All points of the cylinder are at the same distance r from the rotation axis, so that we have I=

X

mi ri2 =

X

mi r 2 = (

i

X

mi )r2 = mr2 .

i

From (??) and (??) we find Fs t. mr The velocity and angular velocity stop changing when sliding stops. The condition of the rolling ω(t) =

40

Physics 1. Mechanics

Problems

without sliding is V = ωr which gives v−

Fs Fs Fs t= tr = t, m mr m

and therefore, in the moment when the velocity stops changing Fs t = v/2. mr Substituting into (??) we have Vf inal = v/2. Problem 10.8. A rod of the mass M and length l can rotate in the vertical plane around the axis which passes through the point at a < l/2 from its upper end. A bullet of the mass m  M v with the horizontally directed velocity v strikes the rod at the upper end and remains in it. What is m, v

the rotation angle of the rod ? Solution. Angular momentum is conserved during the collision, which means Iω = mva → ω = mva/I, E ! mc2

where ω is the angular velocity of the rotation just after the collision, and I is the momentλof inertia. After the collision the energy is conserved, that is, the initial kinetic energy goes into the potential energy as the bar rotates to the angle θ and its center-of-mass rises. Therefore, for the maximum angle we have Iω 2 /2 = M gL(1 − cos θ) → cos θµ= 1 − Iω 2 /2M gL. "m"M M and center-of-mass of the mbar (we neglect the where L = l/2 − a is the distanceµ between the axis mass of the bullet). R positive direction downwards. Calculation of I: Let us choose coordinate origin in the axis, and the

41

Physics 1. Mechanics

Problems

Then Z

l−a

I=

x2 (M/l)dx =

−a

M [(l − a)3 + a3 ] 3l

Combine all calculations. Problem 10.9. An electron moving with the energy E  mc2 toward the coordinate origin emits a photon with the wavelength λ forward. What wavelength measures a non-moving observer in the coordinate origin ? Solution. We shall use directly the expression for the Doppler effect 1 λ0 = . λ γ(1 − v cos θ/c) In our case the source is moving towards the receptor, so that θ = 180◦ and 0

λ = λ

s

1 − v/c . 1 + v/c

We need velocity which can be found as follows: γ = E/mc2 and v = c

p 1 − 1/γ 2 .

Problem 10.10. Potential energy (2D) is given by U = A cos ϕ/ρ2 . A particle of the mass m is in the point r = (a, a) and its velocity is v ⊥ r. Find the normal acceleration. Solution. By definition an ⊥ v. Since r ⊥ v we have an k r. Thus, the magnitude an = |a · rˆ|. √ √ ˆ + y)/ ˆ ˆ + ay)/ ˆ 2a2 = (x 2. Express the potential energy in Cartesian a) The long way. rˆ = (ax coordinates x = r cos ϕ, y = sin ϕ: U = Ax(x2 + y 2 )−3/2 . Then ∂U ∂U ˆ− x yˆ ∂x ∂y     A 3Ax2 3Axy ˆ+ =− 3 − 5 x yˆ r r r5 A(3 cos2 ϕ − 1) 3A sin ϕ cos ϕ ˆ+ ˆ = x y. 3 r r3

F =−

√ Now a = F /m, r = a 2, and ϕ = 45◦ , so that an = A/2ma3 . ˆ so that an = |(∂U/∂r)|/m = |2A cos ϕ/r3 |/m = b) The short way. F = −(∂U/∂r)ˆ r −(1/r)(∂U/∂ϕ)ϕ, A/2ma3 . Problem 10.11. A satellite of the mass m is moving on an elliptical orbit round the Earth (mass M  m), so that rmax = 2rmin . The satellite energy E < 0 is known. Find the angular momentum. Solution. When r = rmin or r = rmax the radial velocity vanishes r˙ = 0, so that in these points

42

Physics 1. Mechanics

Problems

the energy and momentum conservation give E=

J2 GM m . − 2 2mr r

Thus, r1,2 =

GM m ±

p (GM m)2 − 2|E|J 2 /m 2|E|

(here we used E = −|E| < 0) and p

(GM m)2 − 2|E|J 2 /m p = 2. GM m − (GM m)2 − 2|E|J 2 /m GM m +

Solving this equation we get

√ 2GM m m p J= . 3 |E|

Another way of solving: from the energy expression we have J 2 /2m = r2 E + GM mr = (2r)2 E + GM m(2r), so that r = −GM m/3E and J=

p

2m(r2 E + GM mr)

same as above. Problem 10.12. Two disks of the masses m1 and m2 and radii R1 and R2 are connected to a massless rod which can rotate in the vertical plane. The disk centers are at the distances l1 and l2 from the rotation axis. Find the frequency of small oscillations.

m2 , R2 l2

l1 m1 , R1

R h

M r

43

m2 , R2

Physics 1. Mechanics

Problems

l2 moved to the angle θ  1 and the angular velocity is θ, ˙ the energy is Solution. If the system I θ˙2 mgLθ2 I θ˙2 + mgL(1 − cos θ) = + , E= 2 2 2 where

l1

m1 R12 m2 R22 + m1 l12 ) + ( + m2 l22 ) 2 2 m1 ,relative R1 is the moment of inertia to the axis, and I=(

L=

m1 l1 − m2 l2 m1 + m2

is the distance of the center of mass from the axis. The frequency is ω 2 = mgL/I.

m

R

M

Problem 10.13. To a disk of the mass M and radius R a smaller disk of the mass mv and radius h r r is connected. At what height a horisontal force should be applied in order that the body start (in the very first moment) to roll without sliding. There is no friction.

Solution. Let the force be F . The acceleration of the center of mass is a = F/(M + m). The angular acceleration around the contact point is α = N/I = F h/I, where the moment of inertia is I = (M R2 /2 + M R2 ) + (mr2 /2 + mr2 ) = (3/2)(M R2 + mr2 ). If there is no sliding, the linear and angular acceleration are related by a = αL, where L = (M R + mr)/(M + m) is the distance of the B = (0, B, 0) x−y center of mass from the rotation axis (contact point). Thus, we have F/(M + m) = (F h/I)L, so q>0 m that h = I/L(M + m). Problem 10.14. There is a homogeneous magnetic field B = (0, B, 0) between two 2plates, parallel to the x − y plane. A particle of the mass m and electric charge q > 0 enters the space between the planes through the lower plate, when its velocity is v = (v cos 45◦ , 0, v sin 45◦ ). What is the minimal distance between the plates for which the particle will come back to the lower plate

44

Physics 1. Mechanics

Problems v = (v cos 45◦ , 0, v z

v

x (without touching the upper one) ? Gravity is negligible. Physics 1. Mechanics Solutions 2 Solution. The particle moves on a circle of the radius r which is obtained from F = qvB = mv 2 /r, that is, r = mv/qB, as shown in the figure. v1 = (0.5c, 0.5c, 0)

z

|vrel |

v2 = (−0.5c, 0.5c,

R

v

M

x

√ From the figure it is clear that the distance between the two planes should be L > r + r/ 2. √ From the figure it is clear that the distance between the two planes should be L > r + r/ 2.

Problem 1.6. The easiest way is to rotate the coordinates so that the new x axis will be along v1 , √ √ then the new y axis will be along v2 . In the new coordinates v1 = (0.5 2c, 0, 0), v2 = (0, 0.5 2c, 0).

Problem An observer on (S Earth sees one galaxy moving with the velocity v1 = (0.5c, 0.5c, 0) ! Taking v1 ≡ 10.15. v0 as the velocity of the moving frame ), and v 2 as the velocity of a body we are looking at, we have and another with v2 = (−0.5c, 0.5c, 0). Find their relative velocity |vrel |. √ vx − v0 vx! = −v0 rotate = −0.5 2c,the coordinates so(11) Solution. The easiest way is= to that the new x axis will be along v1 , 1 − vx v0 /c2 √ √ √ v v 0.5 2c y y then the new y axis will 1 = (0.5 2c, 0, 0), v2 = (0, 0.5 2c, 0). vy! = be along2 v=2 . In = the .new coordinates v(12) γ(1 − vx v0 /c ) γ γ Taking v1 ≡ v0 as the velocity of the moving frame (S 0 ), and v2 as the velocity of a body we are where we used vx = (v2 )x = 0, and looking at, we have √ γ = (1 − v12 /c2 )−1/2 =

2.

(13)

√ v −v = −v0 = −0.5 2c, 2 For advanced only ! 1 − vx v0 /c √ 0.5 y y Problem 1.7. Let us consider a small part of 0the mass of the v length dr and crossvsection dA at 2c vy = the density ρ = 3M/4πR = = . the radius r. This part has a mass dm = ρdAdr, where γ(1 − vx v0 /c2 ) 3. Is γis attracted toγ ! 3 ! √ √ Thus, we have vx! = −0.5 2c, vy! = 0.5c, and v ! =0 vx! 2 + vy! 2x= 0.5 0 3c.

vx =

the center by the mass M = ρ(4πr /3) inside the radius r. This force is balanced by the pressure forces. From the inside the pressure pushed it outwards with the force p[r]dA. From the outside the

where used vx = and Thus, we have 2 )force x =p[r0, pressurewe pushes it inward with(v the + dr]dA. p[r]dA − p[r + dr]dA =

GM ! dm GMrρ = dAdr.2 r2 R3

γ = (1 − v1 /c2 )−1/2 =

√ (14) 2.

q √ √ 3 Thus, we have vx0 = −0.5 2c, vy0 = 0.5c, and v 0 = vx0 2 + vy0 2 = 0.5 3c. Problem 10.16. Potential energy is given as U = −A/r4 , A > 0. When the particle (mass m) is in the point (a, 0, 0) its velocity is v = (0, v, 0) and its energy is negative. What is the maximum

45

Physics 1. Mechanics

Problems

distance between the particle and the coordinate origin ? Solution. Central force means that J = mr ×v = const and E = mvr2 /2+J 2 /2mr2 +U = const. ˆ J = mav, E = mv 2 /2 − A/a4 . Since E < 0 we have From the initial conditions we have J = J z, v 2 < 2A/ma4 . In the closest and farthest points vr = 0 so that we have J2 A − 4 =E 2 2mr r where J, m, E < 0, and A are known. The equation is easily solved to give J2 ± r2 = − 4m|E|

s (

A J2 2 ) + 4m|E| |E|

2 dve‘zd z‘ ‘vn /2 physical , y = r0 sin(ωt) , xwhich = r0 cos(ωt) wiwlg The solution with. z−=isatnot (r2 < 0) means itl thatrp there is no minimum radius: the particle reaches the maximum radius, bounces back and falls into r .zilnxepd = 0.

Solution .

Problem 10.17. A particle moves according to x = r0 cos(ωt), y = r0 sin(ωt), z = kt2 /2. Find the normal acceleration. v = ωr0 (− sin(ωt)ˆ x + cos(ωt)ˆ y) + atˆz Solution. 2 a = −ω r0 (cos(ωt)ˆ x + sin(ωt)ˆ y) + aˆz

ˆ| an = |a × v ˆ + cos(ωt)y) ˆ + ktzˆ ! v = ωr0 (− sin(ωt)x ω2 a2 t2 + 2a2 + ω4 r02 ˆ + sin(ωt)y) ˆ + k zˆ = ωr0 a!= −ω r0 (cos(ωt)x ω2 r02 + a2t2

an = |a × vˆ| p ω 2 k 2 t2 + k 2 + ω 4 r02 p = ωr0 ω 2 r02 + k 2 t2

3. dl‘y

ly zeiexicz ‘vn . U (x) = A(x − a)2(x − b)2 i’’r dpezp zil‘ivphet dibxp‘

Problem 10.18. Potential energy is U xzei) (x) =lwyn A(x −ieeiy a)2 (x b)2 , Azecepz > 0. Find the frequency of .(cg‘n cil−zephw small oscillations. Solution . Equilibrium: dU/dx = 0=→ x =x a=or to find thatthat U = A(b − a)2 X 2 Solution. Equilibrium: dU/dx 0→ ax or−xb.− It b. isIteasy is easy to find 2 2 2 2 U = A(b − a) X where X = x − a or X = x2 − b. Thus, ω2 = 2A(b − a) /m. where X = x − a or X = x − b. Thus, ω = 2A(b − a) /m.

Problem 10.19. Three identical cylinders are in equilibrium 4. (see figure). Find the minimum dl‘y friction between cylinders. mcwn z‘coefficient ‘va .xeiva d‘xpythe itk lwyn ieeiya mi‘vnd midf mililb dyely .mililbd oia ilnipind jekigd

Solution. Let the normal force acting on the lower cylinder from the upper is N , and the friction Solution .

Let the normal force acting on the lower cylinder from the upper is

N , and the friction force between the upper and lower cylinders is fs , fs /N ≤ µ. From the torque balance on the lower cylinder the friction force between the lower

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Physics 1. Mechanics

Problems

force between the upper and lower cylinders is fs , fs /N ≤ µ. From the torque balance on the lower cylinder the friction force between the lower cylinder and the table is fs . The force balance on the lower cylinder in the horisontal direction is fs + fs cos 30◦ = N cos 60◦ → µ ≥

fs √ = 3/3 N

Problem 10.20. A body is built of two small identical balls connected with a massless rod of the length l. Initially the body is at rest. Another identical ball, moving with the velocity v, collides elastically with one of these two (see figure). Find the angular velocity of the body rotation after the collision.

v

Solution. Elastic collision means that the momentum, angular momentum, and energy are N yig‘ ote‘a (dgepn zkxrna) a, b, after c zecin daiza6. conserved. Let mibltzn the velocity of the third ball thezlra collision be v 0dl‘y (it will be in the same direction v zexidn rpd velocity dtevd cecni xtqn) miwiwlgd zetitv as the initial or in(gtp the zcigil opposite direction - in the last efi‘ case.miwiwlg v 0 < 0). Let the velocity of the a rlv jxe‘l center of the rod be V and the angular velocity of the rotation?around this center (counterclockwise) be ω. Momentum conservation gives Solution . a! = a/γ, V ! = V /γ → n! = nγ mv = mv 0 + 2mV Angular momentum conservation (relative to the collision point): 0 = 2mV (l/2) − 2m(l/2)2 ω Energy conservation gives mv 0 2 2mV 2 2m(l/2)2 ω 2 mv 2 = + + 2 2 2 2 Solving these equations we obtain ω = v/l. Problem 10.21. In a box of the size a × b × c (in its rest frame) N particles are homogeneously distributed. Find the particle density as viewed by an observer moving with the velocity v along a. Solution. a0 = a/γ, V 0 = V /γ → n0 = nγ

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