Rotational inertia and torque of a frisbee

[4t0mic]

Homework Statement

A 103 g Frisbee is 16 cm 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 730 rpm.

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

$$I = \frac{1}{2}MR^{2}$$
$$I = MR^{2}$$
$$\tau = I\alpha$$

The Attempt at a Solution

I already solved for the rotational inertia from the combination of $$I =\frac{1}{2}MR^{2}$$ and $$I = MR^{2}$$. But I can't seem to get how to get the torque from $$\tau= I\alpha$$. I know what I (rotational inertia) but I don't know how to get $$\alpha$$. Help me figure it out! Thanks.

 

[tiny-tim]

Welcome to PF!

With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 730 rpm.
…
… but I don't know how to get $$\alpha$$. Help me figure it out! Thanks.

Hi 4t0mic! Welcome to PF!

(have an alpha: α and an omega: ω )

You know that after a quarter revolution, ω = 730 rpm …

so you have a "distance" and an initial and final "speed" …

so use one of the usual constant acceleration equations

 

[4t0mic]

Hello Tiny Tim.

Oh, I see. So with the constant acceleration equation:
ωf² = ωi² + 2αθ
I get:
α = ωf² - ωi² / 2θ
Giving me:
τ = Iα
τ = I (ωf² - ωi² / 2θ)

Given I = 4.9x10^-4 kg*m² (my answer from part A) and
730 rpm*(1m/60s)=12.1667rps

τ = (4.9x10^-4 kg*m²)([12.1667rps]² - 0 / 2*0.25 rev) = 0.145 kg*m²*r/s²

How do I convert kg*m²*r/s to the SI units for torque, which is N*m? I know N is kg*m/s²... overall I'm stuck.

 

[4t0mic]

Or maybe the question is... how to convert revolutions (r) into SI units?

 

[tiny-tim]

Hello 4t0mic!

How do I convert kg*m²*r/s to the SI units for torque, which is N*m? I know N is kg*m/s²... overall I'm stuck.

Or maybe the question is... how to convert revolutions (r) into SI units?

Exactly!

The great thing about SI units is that if all your inputs are in them, then your result is also …

you don't need to both about what a Newton is (if you break it down) … any force in SI units will be in Newtons automatically!

So, as you say, the only question is... how to convert revolutions (r) into SI units …

well, a revolution is just a dimensionless number, so convert it to radians and you're on a home run