# Inertia and Angular Speed

1. Oct 31, 2013

### Lamebert

1. The problem statement, all variables and given/known data

A block of mass m = 4 kg hangs from a rope that is wrapped around a disk of mass m and radius R1 = 27 cm. This disk is glued onto another disk of again the same mass m and radius R2 = 66 cm. The two disks rotate on a ﬁxed axle without friction. If the block is released at a height 1.7 m above the ground, what is the angular speed of the two disk system just before the block hits the ground. Answer in units of rad/s

2. Relevant equations

Ugrav = mgh

KErot = 1/2 Iω2

Idisk = 1/2mr2

3. The attempt at a solution

Using the work-energy theorem, the work done on the disks by the block is equal to the final kinetic energy of the block, which is equal to the initial gravitational potential energy of the Earth on the block:

ΔKblock = ΔUgrav, block = ΔKrot, disks

Knowing that the final potential energy of the system is approaching zero;

ΔUgrav, block = mgh

So far we have:

mgh = ΔKrot, disks

Knowing also that the initial rotational kinetic energy of the disks is zero, and with equation I provided, we know that

ΔKrot, disks = (1/2)Iω2

Where:

I = (1/2)mr12 + (1/2)mr22

The final equation received would be:

mgh = (1/2)((1/2)mr12 + (1/2)mr222

cancelling for m:

gh = (1/2)((1/2)r12 + (1/2)r222

continuing the move terms over to solve for omega:

4gh = (r12 + r222

4 * (9.8) * (1.7) = [(.27)2 + (.66)2] * ω2

Solving for ω:

131.05 = ω^2
ω = 11.45

This is incorrect though :(

2. Oct 31, 2013

### Staff: Mentor

The block gains some KE as it falls.

3. Oct 31, 2013

### Lamebert

Yep.

4. Oct 31, 2013

### Staff: Mentor

The block loses PE. The disks and the block gain KE.