SOLUTIONS TO CONCEPTS
circular motion;;
CHAPTER 7 1.
Distance between Earth & Moon 5 8 r = 3.85 × 10 km = 3.85 × 10 m 6 T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 sec v=
2r 2 3.14 3.85 10 8 = = 1025.42m/sec T 2.36 10 6
v2 (1025.42)2 2 –3 2 = = 0.00273m/sec = 2.73 × 10 m/sec r 3.85 10 8 Diameter of earth = 12800km 5 Radius R = 6400km = 64 × 10 m a=
2.
V=
2 3.14 64 10 5 2R = m/sec = 465.185 T 24 3600
V2 ( 46.5185 )2 2 = = 0.0338m/sec R 64 10 5 V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a=
3.
v2 22 2 = = 4cm/sec r 1 b) Tangential acceleration at t = 1sec. dv d = (2t ) = 2cm/sec2 a= dt dt c) Magnitude of acceleration at t = 1sec
a=
2 4 2 22 = 20 cm/sec Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m
a=
4.
Horizontal force needed is 5.
in the diagram R cos = mg
mv 2 150 (10 )2 150 100 = = = 500N r 30 30
..(i)
2
mv ..(ii) r Dividing equation (i) with equation (ii) R sin =
Tan =
2
mv2/R
2
mv v = rmg rg
v = 36km/hr = 10m/sec, Tan =
R
mg
r = 30m
100 v2 = = (1/3) 30 10 rg –1
6.
= tan (1/3) Radius of Park = r = 10m speed of vehicle = 18km/hr = 5 m/sec
v2 rg v2 25 –1 –1 rg = tan 100 = tan (1/4)
Angle of banking tan = = tan
–1
7.1
Chapter 7 7.
The road is horizontal (no banking) R
mv 2 = N R and N = mg mv 2 = mg v = 5m/sec, R 25 25 = g = = 0.25 10 100 Angle of banking = = 30° Radius = r = 50m
So
8.
tan =
1 3
v= 9.
mv2/R
g
R = 10m mg
v2 v2 tan 30° = rg rg =
rg 50 10 v2 2 v = = rg 3 3 500 3
= 17m/sec.
Electron revolves around the proton in a circle having proton at the centre. Centripetal force is provided by coulomb attraction. –11 –3 r = 5.3 t 10 m m = mass of electron = 9.1 × 10 kg. –19 charge of electron = 1.6 × 10 c.
mv 2 kq2 23.04 q2 9 10 9 1.6 1.6 10 38 2 = k 2 v = = = 1013 r rm 48.23 r 5.3 10 11 9.1 10 31 2 13 12 v = 0.477 × 10 = 4.7 × 10 6
v = 4.7 1012 = 2.2 × 10 m/sec 10. At the highest point of a vertical circle
mv 2 = mg R 2
v = Rg v =
Rg
11. A celling fan has a diameter = 120cm. Radius = r = 60cm = 0/6m Mass of particle on the outer end of a blade is 1g. n = 1500 rev/min = 25 rev/sec = 2 n = 2 ×25 = 157.14 2 Force of the particle on the blade = Mr = (0.001) × 0.6 × (157.14) = 14.8N The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle also exerts a force of 14.8N on the blade along its surface. 1 12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 33 rpm. 3 1 100 n = 33 rpm = rps 3 3 60 100 10 = 2 n = 2 × = rad/sec 180 9 r = 10cm =0.1m, g = 10m/sec2
10 0.1 r 2 9 mg mr = 10 g
2
2
2 81 7.2
Chapter 7 13. A pendulum is suspended from the ceiling of a car taking a turn 2 r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
mv 2 T sin = r T cos = mg
From the figure
mv2/R
..(i) mg
..(ii)
2 sin mv 2 v2 –1 v = tan = = tan rg cos rmg rg 100 –1 –1 = tan 10 10 = tan (1) = 45°
14. At the lowest pt. T
mv 2 r Here m = 100g = 1/10 kg,
T = mg +
mv T = mg + r
2
(1.4) 1 = 9 .8 10 10
r = 1m,
v = 1.4 m/sec
mg
2
= 0.98 + 0.196 = 1.176 = 1.2 N
mv2 /r
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian. m = 100g = 0.1kg, r = 1m, v = 1.4m/sec. From the diagram,
mv R
T – mg cos =
mg sin
mv 2 T= + mg cos R T=
T
2
mg cos
2 0.1 (1.4)2 (0.1) 9.8 1 1 2
(.2)2 T = 0.196 + 9.8 × 1 2
( cos = 1
2 for small ) 2
T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N 1.16 N 16. At the extreme position, velocity of the pendulum is zero. So there is no centrifugal force. So T = mg cos o 17. a) Net force on the spring balance. 2 R = mg – m r So, fraction less than the true weight (3mg) is
T
mg sin
mg cos
2
2 mg (mg m2r ) 2 6400 10 3 –3 = = = 3.5 × 10 mg g 10 24 3600
=
b) When the balance reading is half the true weight, 2
mg (mg m r ) = 1/2 mg 2
r = g/2
g 10 rad/sec 2r 2 6400 10 3
Duration of the day is T=
2 2 8000 2 6400 10 3 64 10 6 = 2 sec = 2 sec = hr = 2hr 7 3600 9 .8 49
7.3
R mg 2
m /R
Chapter 7 18. Given, v = 36km/hr = 10m/s, r = 20m, = 0.4 The road is banked with an angle, v2 1 –1 –1 100 –1 = tan rg = tan = tan 2 or tan = 0.5 20 10 When the car travels at max. speed so that it slips upward, R1 acts downward as shown in Fig.1 So, R1 – mg cos –
tan = rg 1 tan
V1 =
mv1 /r
R1
mg
..(i)
R2
2
mv 1 cos = 0 r Solving the equation we get,
And R1 + mg sin –
R1
2
mv 1 sin = 0 r
2
2
mv2 /r
..(ii)
0 .1 = 4.082 m/s = 14.7 km/hr 20 10 1 .2
So, the possible speeds are between 14.7 km/hr and 54km/hr. 19. R = radius of the bridge L = total length of the over bridge a) At the highest pt.
mv 2 2 v = Rg v = R 1 b) Given, v = Rg 2
mg =
R2
mg
mv2/R
Rg mg 2
2= L/R mv2/R
mv 2 suppose it loses contact at B. So, at B, mg cos = R 2 v = Rg cos 2 Rv Rg = Rg cos cos = 1/2 = 60° = /3 2 = Rg cos 2
mg 2 2= L/R
R = ℓ = r = r 3 R from highest point 3 c) Let the uniform speed on the bridge be v. So, it will lose contact at distance
The chances of losing contact is maximum at the end of the bridge for which =
mv2/R
L . 2R
mv 2 = mg cos v = R
L gR cos 2R 20. Since the motion is nonuniform, the acceleration has both radial & tangential component So,
2
2= L/R
m
2
v r dv at = =a dt
ar =
mg
mv2/R
mv2/R
2
Resultant magnitude =
v2 a2 r
m dv/dt
2
v2 Now N = m a 2 mg = m r 4
2 2
2
2
2 2
2
2
v2 a 2 2g2 = r 2 1/4
v = ( g – a ) r v = [( g – a ) r ]
7.4
v4 2 r2 a
m
N
Chapter 7 21. a) When the ruler makes uniform circular motion in the horizontal plane, (fig–a) mg = mL
mg
g L b) When the ruler makes uniformly accelerated circular motion,(fig–b)
mg =
2
2
(m2 L ) (mL )
2
4 2
2 g2 + = 2 = L2 2
g 2 2 L
mg
12L
L 1/ 4
R
(Fig–a) m22L
mg
mL
(When viewed from top)
(Fig–b)
22. Radius of the curves = 100m Weight = 100kg Velocity = 18km/hr = 5m/sec a) at B mg –
mv 2 100 25 = N N = (100 × 10) – = 1000 – 25 = 975N R 100
mv 2 = 1000 + 25 = 1025 N A R b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero. 1 = 707N At ‘C’, mg sin = F F = 1000 × 2 At d, N = mg +
B
C
E D
mv2/R B
mv 2 mv 2 c) (i) Before ‘C’ mg cos – N = N = mg cos – = 707 – 25 = 683N R R
mg N
mv 2 mv 2 N= + mg cos = 25 + 707 = 732N R R d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where N is minimum) Now, N = mg sin × 682 = 707 So, = 1.037 F m2r 23. d = 3m R = 1.5m R = distance from the centre to one of the kids N = 20 rev per min = 20/60 = 1/3 rev per sec = 2r = 2/3 15kg 15kg m = 15kg (ii) N – mg cos =
mg ( 2 ) 2 2 2 = 5 × (0.5) × 4 = 10 9 2 Frictional force on one of the kids is 10 24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downward. Here, r = R sin From FBD –1 2 R1 – mg cos – m (R sin ) sin = 0 ..(i) [because r = R sin ] 2 and R1 mg sin – m1 (R sin ) cos = 0 ..(ii) Substituting the value of R1 from Eq (i) in Eq(ii), it can be found out that 2
Frictional force F = mr = 15 × (1.5) ×
1/ 2
g(sin cos ) 1 = R sin (cos sin ) Again, for minimum speed, the frictional force R2 acts upward. From FBD–2, it can be proved R1 that,
R2
m12 r (FBD – 1)
7.5
R1
m22 r (FBD – 2)
R2
Chapter 7 1/ 2
g(sin cos ) 2 = R sin (cos sin ) the range of speed is between 1 and 2 25. Particle is projected with speed ‘u’ at an angle . At the highest pt. the vertical component of velocity is ‘0’ So, at that point, velocity = u cos mv2/r u sin 2 2 centripetal force = m u cos r u cos At highest pt.
mg
mv 2 u2 cos 2 mg = r= r g 26. Let ‘u’ the velocity at the pt where it makes an angle /2 with horizontal. The horizontal component remains unchanged mv2/ u cos ...(i) So, v cos /2 = cos v = cos 2 mg From figure cos
/ mgcos/2
mv 2 v2 r= r g cos / 2 putting the value of ‘v’ from equn(i)
mg cos (/2) =
u 2 cos 2 g cos 3 ( / 2) 27. A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius ‘R’ Friction coefficient between wall & the block is . r=
a) Normal reaction by the wall on the block is = b) Frictional force by wall =
mv 2 R
mv 2 R
mv 2 v 2 c) = ma a = – (Deceleration) R R 2
dv dv v = v =– dt ds R
d) Now, s=
R
ds = –
m
R
R dv v
In V + c
mv2/R
At s = 0, v = v0 Therefore, c = so, s =
R In V0
mv 2/R
R v v –s/R In =e v0 v0 –2
For, one rotation s = 2R, so v = v0e 28. The cabin rotates with angular velocity & radius R 2 The particle experiences a force mR . 2 The component of mR along the groove provides the required force to the particle to move along AB. 2 2 mR cos = ma a = R cos B length of groove = L A 2 2 2 L = ut + ½ at L = ½ R cos t 2 2
t =
2L 2L =t= 1 R2 cos R2 cos
R
7.6
mv /R
Chapter 7 29. v = Velocity of car = 36km/hr = 10 m/s r = Radius of circular path = 50m m = mass of small body = 100g = 0.1kg. = Friction coefficient between plate & body = 0.58 a) The normal contact force exerted by the plate on the block
mv 2 0.1 100 = = 0.2N r 50 b) The plate is turned so the angle between the normal to the plate & the radius of the road slowly increases N=
N=
mv 2 cos r
..(i)
mv 2 sin ..(ii) r Putting value of N from (i)
N=
mv 2 mv 2 –1 –1 cos = sin = tan = tan = tan (0.58) = 30° r r 30. Let the bigger mass accelerates towards right with ‘a’. From the free body diagrams, …(i) T – ma – mR = 0 2 T + 2ma – 2m R = 0 …(ii) 2 Eq (i) – Eq (ii) 3ma = m R
m2R 3 2 Substituting the value of a in Equation (i), we get T = 4/3 m R. a=
****
7.7
m 2m
R
ma
T
2ma
T
a
m2R 2m2R