# Rotational Inertia assistance

1. May 11, 2005

### Bookw0rm

Hello,

After a few hours of searching for an active physics message board, I was so happy to find this one and I am hoping for some assistance

I am not looking for homework answers, I have already completed this assignment to the best of my knowledge. I got my paper back today, and I am hoping to get help with some of the problems I got wrong. (textbook does not help me much here, these are teacher-created problems)

Okay, here we go

1) You have two bowling balls. One is 7 kg and the other is 10 kg. Which bowling ball has the greatest/least amout of rotational inertia and why?

My answer was that rotational inertia is found by (mass)(radius)squared. I reasoned that because both bowling balls have equal radius, the ball with the greater mass (10kg) would have a greater rotational inertia. My answer was marked completely wrong, however. :grumpy:

2) You have an upright, rotating pillar. Attached to the top of the pillar is a string and attached to the bottom of the string is a mass. As the pillar rotates faster, the mass becomes more and more horizontal.
1) Why does the mass travel in a circle?
2) Why does the mass gradually become more horizontal?
3) If the string were to break, which path would the mass take?

This is the one I had the most trouble with. I said that the mass travels in a circle because it has rotational inertia, and does not "want" to change its path. I said that it becomes more and more horizontal because the further from the axis the mass goes, the faster it has to move and the mass will speed up to match the speed of the spinning pillar. I said that if the string were to break, the mass would follow its original path (again because of rotational inertia). I am not sure if my answers were completely wrong or party right, or if the instructor wanted more.

3) You have a 10 m plank, a 100 kg boulder, and a 73 kg person. Ignoring the mass of the plank, where should you place the boulder so that the person may stand on the far end of the plank, with the plank extended 4 m over a cliff (so there are 6 feet on land)?

I used the formula (Fd)cw = (Fd)ccw here, and basicly made a mess of things :yuck: Guidance would be mucho appreciated here!

4) Again with the plank and boulder....Now the plank is 5 m long, and still extended 4 m over a cliff. What size mass do you need on the landside of the plank to keep the person (who is still at the far end over the cliff) from falling?

I said that the mass should be equal to the mass of the person. Wrong. I'm thinking now that I need to use (mass-system)(radius-centerofmass) = (mass1)(radius1) + (mass2)(radius2) or something...

Okay, thats it guys. ANY HELP OR INPUT is very appreciated, these problems are driving me nuts. Anyone that takes the time to work though this and reply is awesome, thank you so so much. Please, if you do reply, tell me which number you are referring to! Thanks!

Last edited: May 11, 2005
2. May 11, 2005

### OlderDan

One at a time will have to do. I'll take breaks in between to look at other things. For this one you are off to a good start. The precise quantity that measures rotational inertia is called moment of inertia. Your problem was not specific about the axis of rotation, and that is important, but if it is the same axis of rotation for both balls and they have the same radius then the ball with greater mass will indeed have a greater moment of inertia. Have you stated the problem correctly? If you have, your answer is correct.

3. May 11, 2005

### OlderDan

It is true that the mass at the end of the string rotating about the axis of the pillar has rotational inertia, but that is not why is is going in a circle. Rotational inertia does not mean things have a tendence to rotate on their own. To maintain rotation, there always needs to be a force toward the center of rotation. Rotational inertia is a measure of the restance to a change in the rate of rotation. The mass is going in a circle because the string is providing a horizontal component of force as well as a vertical component (that equals the weight of the mass). The horizontal component is continually accelertating the mass toward the center of rotation, causing it to travel its circular path. This is called cetripetal (toward the center) acceleration. The natural tendency of the mass is to go in a stright line. It goes in a circle because the string forces it to go in a circle.

The faster the pillar rotates, the faster the mass will move. While the speed of rotation is increasing, the string will not be pointed outward from the center of the pillar. It will be at an angle such that there is a component causing the mass to accelerate in a direction more or less tangent to the circular path.

I think you mean to say that the string becomes more horizontal. As the mass speeds up, more force is required to maintain circular motion Centripetal force is proportional to the velocity squared. At constant speed, the ratio of gravitational force to centripetal force determines the angle of the string (vector addition). As speed increases, the horizontal component of the string tension must increase, becoming larger in proportion to the weight of the mass. The string has to become more horizontal as the centripetal force increases.

If the string were to break, the centripetal force would vanish, and so would the force counteracting gravity. The mass would become a projectile, moving horizontally in a straight line at constant speed while accelerating vertically toward the floor.

4. May 11, 2005

### OlderDan

You have the right formula, you just need to apply it. For the first one, the critical balance will be achieved when the plank just starts to raise the boulder. So take the fulcrum to be the edge of the cliff, call db the distance to the boulder and dp the distance to the person, Wb the weight of the boulder and Wp the weight of the person. Then
Wbdb = Wpdp
The gravitational constant g will divide out, and you can work with masses
Mbdb = Mpdp
but you also know that
db = dp = 10m
You now have 2 equations for the 2 unknown distances. Solve them.

The last one is the same exact problem with the numbers changed. Do it the same way.

5. May 12, 2005

### Bookw0rm

THANK YOU! For responding so quickly

You have explained things so well, and I do understand what you are saying. I had missed several key lectures in this unit prior to the test, and had been relying on notes from friends. I brushed past concepts and memorized formulas, but had not know how to apply them, or more importantly, WHY. Thank you again!

I especially appreciate your explanation of the rotating pillar questions, I had been mixing things up there. You are great!