# Time Dependent Torque/Merry Go Round

1. Mar 6, 2005

### twiztidmxcn

So, you have an uniform merry go round with solid disc radius R and mass M1. It is set spinning with initial angular velocity of omega naut. If it stops after time t1, find magnitude of constant frictional torque that slows it down.

M1 = 243 kg, R = 2.3m, omega nau = 2.5 rad/s, t1 = 30.1 s

So i found frictional torque to be 53.38N. The next part states that you exert a time dependent force tangent to the merry go round as run along side it. This is given by equation F(t) = Fo*e^(-bt) where Fo is F naut and is a constant and b is a constant. You exert the force for time t2 and then jump onto edge of the merry go round and because you move with same speed as a point on the edge of the merry go round, when you jump onto the edge you dont affect its angular velocity. If frictional torque is same throughout entire process, find angular velocity after stop pushing.

Fo = 402N, b = 0.18 s^-1, t2 = 7.8s, your mass is 87kg. So i found omega = 5.381 rad/s.

What was my speed after jump on the merry go round? v = 12.38m/s

Now, here's where the problem is. How many revolutions do you make on the merry go round before it stops? Your mass is 87kg. The frictional torque is the same throughout entire process and when you jump onto the edge of the merry go round, you change its moment of inertia.

I know omega, frictional torque and moment of inertia and that I can find alpha somehow. But how do I find alpha from knowing that information, and how can I find the time until it stops and how many revolutions there are?

I am completely lost at this point, I have a ton of numbers and equations but cannot find something that gives me the right answer. Any help would be appreciated.

2. Mar 7, 2005

### Staff: Mentor

By applying Newton's 2nd law for rotational motion: $\tau = I \alpha$.
Use the kinematic equations for uniformly accelerated motion. You know the initial angular speed and the angular acceleration: Use the rotational analog of $v_f^2 = v_0^2 + 2as$.