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 ✗

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

 

[Nickie]

I would first like to commend you on your attempt to solve this problem using the work-energy theorem and the equations for gravitational potential energy and rotational kinetic energy. However, I would like to point out a few errors in your solution.

Firstly, in the equation for rotational kinetic energy, the moment of inertia (I) should be calculated for the entire system, not just for one disk. In this case, the moment of inertia for the two disks can be calculated as:

I = 2 * (1/2)mr^2 = mr^2

Secondly, in the equation for gravitational potential energy, the height (h) should be measured from the center of mass of the block to the ground, not from the top of the block. In this case, the height can be calculated as:

h = (1.7 + R1 + R2) - R2 = 1.7 + R1

Finally, when substituting the values into the equation, make sure to use the correct units. In this case, the units for mass (m) should be in kilograms, and the units for radius (r) should be in meters.

Correcting these errors, the final equation would be:

mgh = (1/2)(mr^2)ω^2

Substituting the values and solving for ω, we get:

ω = √(2gh/r^2) = √[(2 * 9.8 * 1.7)/(0.27^2 + 0.66^2)] = 4.76 rad/s

Therefore, the angular speed of the two disk system just before the block hits the ground is 4.76 rad/s.