How Is Angular Speed Calculated in a Dual Disk System with a Falling Mass?

[Lamebert]

Homework Statement

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

Homework Equations

U_grav = mgh

KE_rot = 1/2 Iω²

I_disk = 1/2mr²

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:

ΔK_block = ΔU_{grav, block} = ΔK_{rot, disks}

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

ΔU_{grav, block} = mgh

So far we have:

mgh = ΔK_{rot, disks}

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

ΔK_{rot, 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)mr₂²)ω²

cancelling for m:

gh = (1/2)((1/2)r₁² + (1/2)r₂²)ω²

continuing the move terms over to solve for omega:

4gh = (r₁² + r₂²)ω²

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

Solving for ω:

131.05 = ω^2
ω = 11.45

This is incorrect though :(

 

[NascentOxygen]

The block gains some KE as it falls.

 

[Lamebert]

The block gains some KE as it falls.

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:

ΔK_block = ΔU_{grav, block} = ΔK_{rot, disks}

Yep.

 

[NascentOxygen]

mgh = ΔKrot, disks ✗[/size][/color]

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