Lecture 16

Example: The Ballistic Pendulum A ballistic pendulum is a device used to measure the speed of a projectile. As a result of the collision the pendulum-projectile system start swinging. Find the relation between the speed of the projectile and the maximum height the pendulum reaches.

M before

m collision

after

Part I Collision

Conservation momentum only

Part II Swinging

Conservation energy only Note: because of gravity Fext non zero. P is not conserved

What happens if the pendulum is made of an empty box, and the projectile comes out with ½ of its initial velocity? How high will the block swing? In this case you ‘ll need to write a separate eq. for the final momentum of the bullet and block

In this case you only need the conservation of energy for the block.

Systems of variable mass Beginning

Tot. system

u

v

v + dv

dM M

M+dM

dP = (M+dM) (v+dv) – Mv – dM u = Mv +dM v + dM dv – Mv – dM u + Mdv

Σ Fext = dP = v dM - u dM + dv dM + M dv dt dt dt dt dt Σ Fext = v dM - u dM + M dv = (v – u) dM + M dv dt dt dt dt dt Σ Fext= M dv -(u – v) dM dt

dt

vrel of dM with respect to M

dM M dv = Σ Fext + vrel dt

dt

Example of Rocket Find the velocity of a rocket of mass M, after it had emitted some gas. Assume that the fuel is burnt at constant rate, R.

v

vrel = velocity of exhausting gas

vrel

This term is called the thrust of the rocket. Can be interpreted as an additional force on the rocket due to gas expulsion (in the direction of motion) Projecting this vector equation along the direction of v (tangent to the path)

Example

A rocket of mass mR = 2500 kg is accelerated at 3.0g at take off from the Earth. If the gas can be ejected at a rate of 30kg/s what is its exhaust speed?

Fext + vrel

dM = Ma dt

dM = M (3g) -Mg + vrel dt

vrel ~ 3.3 x 103 m/s

Example

A cart of mass mc = 14000kg is rolling horizontally with vc=4m/s. As it passes by a grain elevator, 2000 gr of grain are suddenly dropped into the car. If d = 500 m is the distance from the elevator to the switch yard, how long does it take to the cart to cover this distance? Note: Momentum is conserved along the x-direction because there are no external force along x.

d

The time for the car to travel is:

∆t = d / vF

We can find vF from conservation of momentum (mc + mg) vF = mcvi + mg (0) ∆t = d (mc + mg) / mc vi

vF = mc / (mc + mg) vi

vF =3.5m/s

∆t = 143 s

Note: Mechanical energy is conserved in thermal energy E0= ½ mc vc2 + Ki grain = 14KJ

Ef =Kf= ½ (mc + mg) vf2 = 98KJ Ki < Kf

Example

A uniform rope of mass M and length L is held with its lower end just touching the surface of a scale. The rope is released and begins to fall with constant velocity. Find the force F on the scale on the rope as L/2 of the rope touch the scale.

Start L

End L/2

L/2

Example: Two blocks of mass 3kg each are on a car. Beginning from rest the driver of the car (neglect his mass) throws the weight horizontally one at a time. The speed of each weight relative to him is vWC. Calculate how his velocity will change after he has thrown the first and the second weight.

Part 1: Throw first weight Initial velocity car Final velocity car relative to car relative to car

car

car

Part 2: Throw second weight

Solve as part 1. In this case you start with a car + weight and the final system is only the car.

Impulse How can we describe it in terms of force?

Let’s look at the collision between two particles

Variable force, max at moment of collision Fmax

Fnet dt

J=

ion ans exp

t2

compr ession

F

t

t1

J is the area under the F(t) curve! Contact begins

Contact ends

Impulse From Newton’s 2nd law:

dp = Fnet dt

dp = Fnet dt

t2

dp = Fnet dt

integrate

p1

t1

Definition

t2

Fnet dt

∆p =

p2

t1

t2

Fnet dt

Impulse = J = t1

[J] = [ N sec]

Measure the change in momentum

Impulse-Momentum Theorem

∆p = J

Tells us that we don’t need to know all the details of F(t) to know the velocity of an object

Case: F = constant or Faverage t2

Fnet dt

J= t1

J = Faverage ∆t

Example:

vf

θ v0

1) Find impulse of the force exerted on a ball of mass m. 2) Assuming the collision lasts for ∆t = 1.5 ms, what is the average force? 3) Find the change in the momentum of the bat.

Center of Mass Show demo in class So far we assumed that any extended body can be approximated as a point that undergoes only translational motion. But in real life any object undergoes both rotation and translation. Is this wrong?

We can always find a point whose motion is only a translation and moves in the same path as a particle: CENTER of MASS (CM)

Newton’s second law for a system of particles The center of mass of a body or a system of bodies moves as though all of the mass were concentrated there and all external forces were applied there.

MaCM = ∑ Fext

Motion of CM

We can treat the translational motion of any body or system of bodies as the motion of a particle! Note: Internal forces cancel each other because of Newton’s third law

How to determine the CM? 1D case System of two masses Definition

xCM =

0

m1x1 + m2x2 m1 + m2 CM

Case I m1 = m2

xCM =

x2

x1

0

m1x1 + m2x2 m1 + m2

x2

x1

=

m (x1 + x2 ) 2m

=

(x1 + x2 ) 2

d =

2

The CM coincides with the geometrical center

Case II m1 > m2

CM

0

m1 (x1 + x2 m2 / m1)

m1x1 + m2x2

xCM =

x2

x1

xCM =

m1 + m2

In the extreme case m2=0 xCM = x1

xCM closer to m1

CM

Case III m1 < m2 xCM =

m1 (1+ m2 / m1)

~ x1

0 m1x1 + m2x2 m1 + m2

x1

x2 xCM closer to m2 In the extreme case m1=0 xCM = x2

n

For a system of n particles: xCM =

M = tot mass system

∑ mixi

i =1

M

CM in more than 1D For each component n

n

rCM =

∑ miri

i =1

M

n

∑ mixi

i =1

xCM =

M

Example : Find the CM of H2O molecule We can find first the CM of the two H atoms and then the CM of (2H) with the O

CM

yCM =

H

i =1

M H

mO = 16u

O

mH = 1u

α = 52.2ο

H

H 2mH

∑ miyi

mO = 16u

O

CM (H2O) 2mH CM

CM for a solid body ∆m 1) Divide it in n infinitesimal masses, dm 2) Calculate the CM for the new system n masses n

rCM =

∑ ∆miri

r dm

M

M

i =1

∆mi

0

rCM =

r dm M

Center of gravity In a similar way we can define the CENTER of GRAVITY as the point where gravity can be cosidered to act. In a system with constant g: CM = CG

Example with CM m1 m2

A wedge of mass m2 and a block of mass m1 are on a balance. When the block is released it slides down on the wedge. Find the reading of the scale