A uniform, solid disk of radius 0.12 m and mass 5 kg is pivoted so that it will rotate around its central axis

Question

A string wrapped around the disk is pulled with a constant force of 20 N. If the disk starts from rest,

a.) what is its kinetic energy after 5 s?
b.) what is the total angle change in theta that the disk turns through 5 s.
c.) show that the work done by the torque, tau * change in theta, equals the kinetic energy.

Answer 1

This is basically a yo-yo on it's side... kind of.

Before we start, we need to know the moment of inertia for the solid disk. For a solid disk, I = ½m R²

For our disk, I = 0.036 kg m²

The disk is not moving up and down, so there is no change in GPE. The disk is only spinning, not moving back and forth, so there is no linear velocity. Thus, the only energy is found in the rotational kinetic energy, or KE = ½ I ω².

So we need to find ω, the angular velocity.

Since F = ma, 20 = 5 * a, and the linear acceleration is 4m/s².
To find the angular acceleration, α, we use the formula

a = r*α

to get α = 33.33 rad/s²

Now we can calculate ω

α = (ωf - ωi)/(tf - ti)

α = (ωf - 0)/(5 - 0)
33.33 = ωf/5
ω = 166.7 rad/s

KE = ½ I ω² = .5 * 0.036 * 166.7² = 500.2 J

That's a lot for one problem!

b) Well, we know α = 33.33 rad/s², and ωi = 0 rad/s and t = 5 s

θ = ½α t² + ωi t = 0.5 * 33.33 * 5² = 416.6 rad

c) The work done by torque is

W = τ θ = (I α) θ = (.036 * 33.33) * 416.6 = 499.87 J

Comparing 500.2 J to 499.87 J and realizing we only need one sig fig.. 500 J seems perfect