A pulley with rotational inertia

[soccer_09]

Homework Statement

A pulley, with a rotational inertia of 2.0 10-3 kg·m2 about its axle and a radius of 20 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t2, where F is in Newtons and t in seconds. The pulley is initially at rest.

(a) At t = 4.0 s what is its rotational acceleration?
1 rad/s2
(b) At t = 4.0 s what is its rotational speed?
2 rad/s

Homework Equations

Torque = I * alpha

Torque = |r||F|sin(theta)

The Attempt at a Solution

I have only tried part a. What I did was since I know the radius = .020 m, F, and I, I rearranged the formulas like so:

alpha = Torque / I

Since it's tangent, sin(90) = 1 therefore Torque = r*F

F at 4s = 6.8 N and multiplying this by the radius .020 m gives a torque of .136 Nm

So now alpha = .136 Nm / .002 kgm^2 = 68 rad/s^2

I plugged that in on webassign and it was wrong so apparently I'm not doing something right. Any guidance is appreciated, thanks.

 

[kuruman]

Your method for calculating the angular acceleration is correct. However 20 cm = 0.20 m not 0.02 m.

 

[soccer_09]

Yep, I had a friend tell me I converted incorrectly. Thanks :)

From there though, my alpha = 680 rad /s^2

The equation for rotational veloctiy is:

omega(t) = omega (initial) + alpha (t)

so with alpha = 680 rad/s^2 and t = 4s that should give me a rotational speed of 2720 rad/s

This, however, is incorrect as well. The initial omega = 0 rad/s because the pulley is initially at rest. What am I doing wrong here??

 

[ideasrule]

The equation omega(t) = omega (initial) + alpha (t) only works for constant angular acceleration. Here, force is F = 0.50t + 0.30t2, so alpha can't possibly be constant. You'll have to use integration to get the speed.

 

[soccer_09]

Thanks :) I integrated from 0 to 4 of F(t) and got 10.4. using the constants I and R, I was able to get the correct answer. Thanks again.