# Rotational Dynamics and Equillibrium HELP

1. Jun 10, 2006

### cukitas2001

Hey guys, i dont know why its always two or three problems on my homework that always stump me. Any help and tips are appreciated

1) A wheel with a weight of 393N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 27.6 rad/s. The radius of the wheel is 0.645m and its moment of inertia about its rotation axis is 0.800MR^2 . Friction does work on the wheel as it rolls up the hill to a stop, at a height of above the bottom of the hill; this work has a magnitude of 3536J.

Part A) Calculate h. Use 9.81m/s^2 for the acceleration due to gravity.

I tried an energy approach and got an expression for h as follows:
h=(2*3536)/(I*omega^2) where I was moment of inertia....i think i messed up in the friction part though. Any advice?

2) A large turntable rotates about a fixed vertical axis, making one revolution in a time of 5.90s. The moment of inertia of the turntable about this axis is 1250kg*m^2 . A child with a mass of 45.0kg, initially standing at the center of the turntable, runs out along a radius.

Part A) What is the angular speed of the turntable when the child is a distance of 2.40m from the center? (Assume that you can treat the child as a particle.)

No idea where to even being on this one.

2. Jun 10, 2006

### Pyrrhus

Well down hill the wheel has both translational and rotational kinetic energy, and when is going to a height h it only has gravitational potential energy. Your expression seems incomplete for A.

Last edited: Jun 10, 2006
3. Jun 10, 2006

### cukitas2001

How can i take into account both translational and rotational values then? btw i figured out 2 so just focusing on 1 now

I had forgotten to take into account frictional work done so i came up with a new expression for h: (.800(393/9.81)*(.645^2)*(27.6^2)+3536)/(2*393)
this gives me a height of 17.4 m, does this seem right?

Last edited: Jun 10, 2006
4. Jun 10, 2006

### Pyrrhus

Well write out the equation

$$\Omega_{gravity} = K_{tras} + K_{rot}$$

$$mgh = \frac{1}{2}mv^{2} + \frac{1}{2}I_{cm}\omega^{2}$$

5. Jun 10, 2006

### cukitas2001

What aobut the fricitonal work of 3536 J ?

6. Jun 10, 2006

### Pyrrhus

Oh didn't notice, yes it has to be added, because it does work on the wheel.

7. Jun 10, 2006

### cukitas2001

ok so im using:

h=(.5*40.1*(17.8^2)+.5*.8*40.1*(.645^2)*(27.6^2)+3536)/393

where 40.1 is the mass 40.1kg of the wheel weighing 393N and 17.8 is the translational velocity using the relation v=r*omega. I'm getting 38.1m but it doenst seem right though...was i supposed to subtract the friction work?

nvm it was minus friction force got the answer

Last edited: Jun 10, 2006
8. Nov 15, 2006

### curiozity

What about the second problem?

I happened to come across this post.
What about the second problem in the orginial post?
How would you do that problem?
I'm just curious...
Anyone willing to teach?
P.S. I'm not a Physics major, so I don't know everything about it (sorry no offense to anyone ) I was thinking about majoring in Physics.
I think I could probably follow it if someone just did the problem like a book example.

Last edited: Nov 15, 2006
9. Nov 16, 2006

### turdferguson

Angular momentum is conserved. With linear momentum, if the mass increases, the velocity increases. For rotation, you use moment of inertia and angular velocity. When the child moves from the center outward, he increases the moment of inertia of the system, so the angular velocity must decrease. Its nice that you can treat him as a point particle because he has no effect on I if the radius is zero