2nd law of rotation, merry go round

[Ready2GoXtr]

Homework Statement

http://ready2goxtr.googlepages.com/problem1052.jpg

A uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of 2.00 cm and a mass of 20.0 grams and is initially at rest. Starting at time t = 0 two forces are to be applied tangentially to the rim as indicated, so at time t = 1.25s the disk has an angular velocity of 250rad/s coutnerclockwise. Force F1 has a magnitude of .100 N. What is the magnitude of F2.

Homework Equations

T(net) = I$$\alpha$$
$$\omega$$ = $$\alpha$$ t <---- Equation 2

The Attempt at a Solution

r = 2.00cm = .2m m = 20.0g = .02kg t = 1.25s $$\omega$$ = 250 rad/s

Using equation 2 I find $$\alpha$$ = 200 rad/ s^2

I did try one way converting $$\alpha$$ into a linear acceleration by multiplying it by r Then said that F1 + F2 = ma

Didnt work =(

 

[cryptoguy]

I did try one way converting $$\alpha$$ into a linear acceleration by multiplying it by r Then said that F1 + F2 = ma

Didnt work =(

I would try to go the other way around, convert everything to rotation terms: you know $$\tau_{net} = I \alpha$$ so $$F_1 r - F_2 r = I \alpha$$.

 

[Doc Al]

Using equation 2 I find $$\alpha$$ = 200 rad/ s^2

Good.

I did try one way converting $$\alpha$$ into a linear acceleration by multiplying it by r Then said that F1 + F2 = ma

Don't convert anything. What net torque is required to produce such an angular acceleration? (What's the rotational inertia of the disk?)

 

[Ready2GoXtr]

How do i find the rotational inertia of the disk, its uniform, so does mr^2 work?


or is it... 1/4 mr^2

 

[Doc Al]

How do i find the rotational inertia of the disk, its uniform, so does mr^2 work?


or is it... 1/4 mr^2

Don't just guess. Look it up!

 

[Ready2GoXtr]

It just gives it to me in x y and z components.

 

[Doc Al]

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

 

[Ready2GoXtr]

I think its 1/2 mr^2 in which ase I = .0004 so I (alpha) = .08


F1r - F2r = .08
F2 = -.3 N

 

[Doc Al]

I think its 1/2 mr^2 in which ase I = .0004

Correct formula, but recheck your arithmetic.