# Rotational equilibrium

1. Apr 26, 2007

### chamonix

1. The problem statement, all variables and given/known data
A 12 kg mass is attached to a cord that is wrapped around a wheel with a radius of 10.0 cm. The acceleration of the mass down the frictionless incline of 37 degrees is measured to be 2.0 m/s^2. Assuming the axle of the wheel to be frictionless, determine:
a). the force in the rope.
b). the moment of inertia of the wheel.
c). the angular speed of the wheel 2.0 s after it begins rotating, starting from rest.

2. Relevant equations
I=1/2mr^2

3. The attempt at a solution
For a). I got 14.44 N by using sin37*2*12
for b). I used I=1/2mr^2 and got .06
for c). I did not understand how to get this answer.

2. Apr 26, 2007

### hage567

Not quite right. Set up an equation with F=ma, and sum up the forces. Remember, the block is accelerating, it's not at rest.
But you don't know the mass of the wheel, 12 kg is the mass of the block. The tension due to the block is causing a torque on the wheel. You must relate the two to find the moment of inertia.
You were given the linear acceleration so you can get the angular acceleration. From that, using rotational kinematics, you can find the angular speed.

3. Jul 6, 2008

### nopistons93

for a) what are the external forces? I said F=ma = (12.0 kg)(cos 37)(2.0m/s) = 19.17 N

b) I said Torque = radius*force = (0.1 m)(12 kg*cos(37)*2.00m/s) = 1.92 N*m

thus solving for in $$\tau$$=I$$\alpha$$ , I calculated 0.0958

c) I've calculated 14.1 m/s from using a=r$$\alpha$$
then I used $$\alpha$$=r$$\omega$$2 and solved from there. does that sound correct?

4. Jul 6, 2008

### nopistons93

sorry for part b I got I=0.0958

5. Jul 7, 2008

### nopistons93

anyone to confirm?

6. Jul 7, 2008

### nopistons93

a)

F= mgsin(theta) - ma

b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha.

c) alpha = r*angular vel^2 , solve for ang vel and your done

7. Jul 8, 2008

### alphysicist

Hi nopistons93,

The formula highlighted in red is not correct.