I am reposting this quesiton because the previous answer I was given for part B

Question

I am reposting this quesiton because the previous answer I was given for part B (0.05 Nm) was incorrect.

A 101 g Frisbee is 33 cm in diameter and has half its mass spread uniformly in the disk and the other half concentrated in the rim.

A) What is the Frisbee's rotational inertia? (in kgm^2)

Answer: 2.1*10^-3

B) With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 720 rpm . What is the magnitude of the torque, assumed constant, that the student applies? (in Nm)

Answer: ???????

Explanation / Answer

Moment of Inertia, I = 1/2 *1/2 m R^2 + m R^2
I =1/4 * m R^2 + 1/2 mR^2
I = 3/4 *mR^2
I = 3/4 * 101/1000 * (0.33/2)^2 Kg m^2
I =  2.1 * 10^-3 Kg m^2

Torque = I *

= 720 * (2*pi)/60 rad/s
= 75.4 rad/s

Quarter a turn flick means,
= 1/4 * (2*pi) rad
= 1.571 rad

we know,
² = ² + 2
Solving for -
= ²/(2*)
= 75.4^2 /(2*1.571)
= 1809.4 rad/s^2

Torque = I *
Torque = 2.1*10^-3 * 1809.4 Nm
Torque = 3.8 Nm

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