Potter's Wheel Angular Acceleration Calculation

[mustang]

Problem 3.
A potter's wheel of radius 0.19m and mas 82.10 kg is freely rotating at 46.0 rev/min. The potter can stop the wheel in 8.4 s by pressing a wet rag against the rim
a. What is the angular acceleration of the wheel? In rad/s^2.
Note: Is the formula to use in this problem t=I*a?Where I is moment of inertia and a = angular acceleration. How would the data in the problem be set up in the formula?

 

[HallsofIvy]

"Is the formula to use in this problem t=I*a?Where I is moment of inertia and a = angular acceleration."
What is t? Normally t means time but that doesn't make sense. I*a = force.

Acceleration is change in speed divided by the time required for the change (the problem says "In rad/s^2"- that's rad/s divided by s). Here you are told that the initial speed is 42 rev/ min (how many revolutions is that per second) and that it slows to 0 in 46 seconds. That's all the information you need.

 

[mustang]

t= torque.
This equation is Newton's second law of rotation. I was wondering if I= 12.5kg*m^2, then to find torque it is force times distance. I was wondering what would be the a to find the force? In addition if it is rev./min for a, do i just leave it in there?

 

[gnome]

I don't see how you're getting that value for I.
For this object
I_cm = (1/2)mr²

But anyway, it's irrelevant here.

As HallsofIvy already told you, you don't need torque or moment of inertia here. You have been given the angular velocity in rev/min, and you are told how long it takes to stop it.

For linear motion:
a_avg = Δv/Δt, right?

Just do the equivalent for angular motion (after converting your given angular velocity to the appropriate units).