# 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?
(b) At t = 4.0 s what is its rotational speed?

## 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.

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

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??

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.

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.