Rotational Inertia and Torque for a Spinning Frisbee

[Nick_L]

Homework Statement

A 120g Frisbee is 28cm in diameter and has about half its mass spread uniformly in a disk, and the other half concentrated in the rim. With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 560rpm.

(a) What is the rotational inertia of the Frisbee?

(b) What is the magnitude of the torque, assumed constant, that the student applies?

Homework Equations

I=1/2*m*r²

I=m*r²

\tau=I\alpha

The Attempt at a Solution

I got the first part of this by using I=1/2*.60kg*.14m² for the disk then adding I=.60kg*.14m² for the rim which gives me 0.001764kg/m²
The part that is messing me up is finding the torque, I think the best way to find the torque is to find the angular acceleration of the frisbee, but I havn't been able to get it. I am pretty sure I need to use the rotational/kinematic equations... Any help would be great.

 

[LowlyPion]

Welcome to PF.

First of all maybe recalculate your I ?

 

[LowlyPion]

So long as you convert angles to radians and ω to radians/sec then you can use the kinematic analogs to motion:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

You are accelerating the frisbee from rest to 560 rpm over 1/4 a turn (according to the problem anyway).

 

[Nick_L]

Thanks LowlyPion, I didn't realize it was only a quarter turn.

 

[LowlyPion]

Be sure and correct your moment of Inertia calculation.

3/2(.60)(.14)² is not .00176

Edit: Wait. I see it should have been .06 kg not as you wrote it. Your answer for I is correct making that change.