Question

A block hangs from a string that runs over a pulley and is attached to a rolling cart. The cart, the block, and the pulley each have the same mass (M) . The cart an pulley roll without friction, and the pulley is a disk with moment of inertia:

Icm=12MR2

The hanging block is released from rest and allowed to evolve freely under the influence of gravity.

Find the speed of the hanging mass, v, after the block has fallen over a distance equal to 5R. Give your answer a a function of M, R, and the gravitational acceleration g, as needed.

Answer 1

Consider pulley,cart and earth as a system.

So, the forces acting on the system include

1.gravitational force which is an internal conservative force for the system.

2.Tension due to string. But total work done by tension is 0 if string is inextensible. So,here, total work done by string=0

3.normal reaction on pulley.

Work done=Fs cos where F and s are magnitudes of force and displacement vector respectively, and is the angle between force and displacement vectors.

In case of normal reaction, Normal force is perpendicular to surface, while displacement is along the surface.So,angle between normal reaction and force is 90 So,work done by normal reaction=Fs cos90=0

So,normal reaction does not do any work

4.Hinge force on pulley. Work done=0 as point of application of hinge force(center of pulley) does not move.

So,work is done only by gravitational force which is an internal conservative force for the given system.

If only internal conservative forces do work,mechanical energy is conserved.So, here, mechanical energy is conserved.

initial situation:

Potential energy=mgh,where m is mass, g is gravitational acceleration, h is height.

Let initial height of block be h and that of pulley and cart be H.

So, initial potential energy=Mgh+MgH+MgH

Kinetic energy of translating objects=1/2mv² where m is mass, v is velocity.

Kinetic energy of rotating objects=1/2I² , where I is moment of inertia, is angular velocity

So, kinetic energy of block=1/2M v²=0 as v=0 m/s( block starts from rest)

kinetic energy of cart=1/2M v²=0 as v=0 m/s( cart starts from rest)

kinetic energy of pulley=1/2*I ²=0 as =0 m/s( pulley starts from rest)

So,initial mechanical energy=kinetic energy+potential energy=Mgh+MgH+MgH

final situation:

Potential energy=mgh,where m is mass, g is gravitational acceleration, h is height.

final height of block = h-5R (R is radius of pulley)as block falls down by 5R and height of pulley and cart = H.(final height = initial height for pulley and cart as they don't move vertically)

So, final potential energy=Mg(h-5R)+MgH+MgH

Kinetic energy of translating objects=1/2mv² where m is mass, v is velocity.

Kinetic energy of rotating objects=1/2I² , where I is moment of inertia, is angular velocity

Let the final velocity of block be v m/s

Then,final velocity of cart is also v m/s as cart and block are connected by inextensible string.

Also,assuming no slipping, angular velocity of pulley =v/R where R is radius of pulley.

So, kinetic energy of block=1/2M v²

kinetic energy of cart=1/2M v²

kinetic energy of pulley=1/2*I ²=1/2(MR²/2)(v/R)² as I=MR²/2 for pulley(disc)

=1/4Mv²

So,final mechanical energy=potential energy+kinetic energy=Mg(h-5R)+MgH+MgH+1/2Mv²+1/2Mv²+1/4Mv²

Using energy conservation,initial mechanical energy=final mechanical energy

=>Mgh+MgH+MgH=Mg(h-5R)+MgH+MgH+1/2Mv²+1/2Mv²+1/4Mv²=>5/4Mv²=5MgR=>v=(4gR)^1/2

So,required velocity of hanging mass=(4gR)^1/2