# 2nd law of rotation, merry go round

1. Jul 11, 2008

1. The problem statement, all variables and given/known data

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.

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

3. 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 =(

2. Jul 11, 2008

### cryptoguy

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

3. Jul 11, 2008

### Staff: Mentor

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

4. Jul 11, 2008

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

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

5. Jul 11, 2008

### Staff: Mentor

Don't just guess. Look it up!

6. Jul 11, 2008

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

7. Jul 11, 2008

### Staff: Mentor

Last edited by a moderator: Apr 23, 2017
8. Jul 11, 2008

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

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

9. Jul 11, 2008

### Staff: Mentor

Correct formula, but recheck your arithmetic.