# Homework Help: Rotational dynamics

1. Nov 16, 2008

### zoner7

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

A solid ball, with radius r = 6.00 centimeters and mass m = 0.400 kg, rolls without slipping on the inside of a fixed horizontal circular pipe with radius R = 0.300 meters. The ball is held at rest with its center lying on the horizontal diameter of the pipe. It is then released and rolls down the side of the pipe, in a plane perpendicular to the length of the pipe. (a) What is the total kinetic energy of the ball when it reaches the bottom of the pipe? Hint: You must take into account the distance between the ball's center of mass and the inner surface of the pipe. (b) What is the speed of the ball's center of mass when it reaches the bottom? (c) What normal force does the pipe exert on the ball when the ball reaches the bottom?

2. Relevant equations

3. The attempt at a solution

Well... I've got ideas, but I really don't have any confidence in any one of them in particular.

I considered finding the arclength of the tube so that we know how the distance that the ball travels. This would enable us to find other values using rotational motion equations.

I also was looking at the kinetic energy and potential energy formulas. Clearly these are necessities.

The hint " Hint: You must take into account the distance between the ball's center of mass and the inner surface of the pipe" frankly confuses me more than anything. I'm not even sure what this means nor do I understand how it is even relevant to the solution.

I think I'm going to go back and reread the chapter. if anyone could offer a little insight while I do this, I would be very appreciative. I just need a starting point.

Thank you :)

2. Nov 18, 2008

### Redbelly98

Staff Emeritus
Yes, that is the key. By how much does the potential energy change when the ball moves from it's starting point to the bottom of the cylinder? (Answer depends on the change in height of the ball's center of mass.)

3. Nov 19, 2008

### zoner7

ooo... so I think I get what you mean... let me work it out

4. Nov 20, 2008

### zoner7

alright. here we go

a)

The distance between the ball and the center of the pipe is .24m (R - r) - we will call this new value h.

KEf here is defined as the kinetic energy of the ball at the bottom of the pipe.
KEf = PEi
so, KEf = mgh
KEf = .941 J

b)
now we need to find the velocity at the bottom of the pipe. we know that the kinetic energy will be divided into both translational and rotational energy.

KEf = 1/2mv^2 (translational) + 1/2IW^2 (rotational)

now.. we want to find the translational velocity, not energy...

Let's substitute in the moment of inertia of a solid ball,
KEf = 1/2mv^2 + (1/2) ((2/5)mr^2) W^2

now we need to eliminate the W so that we have a single variable.
We can do this by multiplying by (v^2/V^2)
KEf = 1/2mv^2 + (1/2) ((2/5)mr^2) W^2 (v^2 / v^2)

Since Wr = v, we can say that KEf = 1/2mv^2 + 1/10 mv^2

Now... I'm quote sure what to do about 2 v's... I'm going to take a guess that both v's refer to the linear velocity and that the v that came from the rotational velocity no longer has anything to do with rotation.

Who can give me a hand here?

5. Nov 20, 2008

### Redbelly98

Staff Emeritus
a) You are correct.

b)
You're right, sort of. v does still have something to do with rotation. But since the ball is rolling without slipping, the two v's are ______ .

6. Nov 20, 2008

### zoner7

... My guess is that both v's now refer to the linear velocity. Mathematically that makes sense, but I do not quite understand it conceptually.

7. Nov 21, 2008

### Redbelly98

Staff Emeritus
Your guess is correct, here is the explanation.

It's easier to think of a car's wheel on a flat road. And, put yourself in the reference frame of the car, so that it is the road that is moving past you at a speed vroad.

In your reference frame, the wheel's center is stationary, but since it is rotating then the outer tire edge is moving at a speed vedge.

The only way for the tire not to slip or skid on the road surface is if those two speeds are equal. If that still doesn't make sense, I could probably come up with a figure to show it better.

8. Nov 21, 2008

### zoner7

I think I understand what you mean. Not positive.. but I think I'm there.

And I'm going to take a good guess that the normal force is simply determined by the mass times centripetal acceleration where the r value is the previous defined variable h, the distance between the balls center of mass and the center of the pipe...

9. Nov 21, 2008

### Redbelly98

Staff Emeritus
It is a good guess, since that h is the radius of the path of the ball's center-of-mass. It would be better to understand why that is so, rather than guessing, but it is correct.

Of course, multiplying centripetal acceleration by mass would give the net force acting on the ball.

10. Nov 21, 2008

### zoner7

I understand why the the ball's canter of mass is h, but I'm still trying to grasp the concept that kinetic energy is divided to create both linear and rotational velocities.

11. Nov 21, 2008

### Redbelly98

Staff Emeritus
Ah, yes. Well, it's really just a mathematical shortcut or convenience that that's the case.

Imagine subdividing the wheel into smaller units ... even to the level of the molecules. To get the total kinetic energy of the object, calculate the kinetic energy of each unit, (1/2)mv2, and then add up these individual kinetic energies.***

Or ... you could calculate just two kinetic energy terms: rotation, and center-of-mass translation. Either way you get the same answer.

Here's a simple example demonstrating this.

Imagine four 1 kg masses located at edge of a rotating, moving circle of radius 1m. Let the masses be symmetric at 90 degree intervals around the circle, and
1. the c-o-m is moving at 1 m/s
and
2. this formation is rotating at 1 rad/s

So
(1/2)mc.o.m.v^2
= (1/2)(4kg)(1m/s)^2
= 2J
and
(1/2)Iω^2
= (1/2)(4kg-m^2)(1/s)^2
= 2J
Total KE = 2J+2J = 4J

Now let's calculate and add up the KE's of the individual masses:
Suppose the c.o.m. is moving to the right, and the rotation is clockwise. This means that:
a mass is stationary when it's at the bottom of the circle
a mass moves at 2 m/s when it's at the top
a mass moves sqrt(2)m/s diagonally when it's at the sides.

The 4 KE's summed together are
(1/2)(1kg)(0m/s) + (1/2)(1kg)(sqrt(2)m/s)^2+ + (1/2)(1kg)(sqrt(2)m/s)^2+ + (1/2)(1kg)(2m/s)^2
= 0J + 1J + 1J + 2J
= 4J, just as before.

I'm sure there is a general proof of this, but it seemed easier to give a simple example.