# Torque/moment of intertia questions

1. Jan 22, 2007

### Koolaid

i have two questions that i need help understanding

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

A majorette takes tow batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.6m long and each ball on the end is 0.50kg. (Ignore the mass of the rods.) What is the moment of inertia if the arrangement is spun around an axis formed by one of the batons?

2a. Relevant equations

I=(1/12)ML^2

3a. The attempt at a solution

Not sure how to apply the equation to the problem. And what does it mean when n the arrangement is spun around an axis?

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

A tether ball is attached to a pole with 4.0m rope. It is circling at 0.20 rev per second. As the rope wraps around the pole it shortens. How long is the rope when the ball is moving at 6m/s

2b. Relevant equations

not sure

but .20 rev/sec = 1.26 rads/sec

3b. The attempt at a solution

not sure what to do.. what equations should i be focusing on?

Last edited: Jan 22, 2007
2. Jan 22, 2007

### PhanthomJay

In problem one, the rod has no mass; you should focus your attention on the moment of inertia of the 'point' masses at the ends of the rods. If you consider one of the rods as the x axis, then the other rod is spinning clockwise or counter clockwise around that rod, as opposed to, for example, the case where both rods are rotating about an axis that is perpendicular to their plane of rotaion. You are looking at the first case. Identify the moment of inertia formula, and determine the distance of each ball mass as measured to the axis of rotation.

3. Jan 22, 2007

### Koolaid

if i do that (.5*.3^2) times to for each ball... so .09 kg-m^2... doest hat make sence?

4. Jan 22, 2007

### PhanthomJay

Makes a lot of sense, great. The other 2 balls don't contribute to the moment of inertia, since there is no distance of those masses to the axis of rotation. Now, page 2. Focus on conservation of angular momentum for the 2nd problem.

5. Jan 22, 2007

### Koolaid

well the conservation of angular momentum is L=Iw(w=omega)

but i can't seem how to fit the givens in or come up with missing pieces like the moment of inertia

6. Jan 22, 2007

### PhanthomJay

yeah, it's L_(initial) =L_(final). You can calculate I_(initial) now in terms of the ball's mass. You are given w_(initial). You are given the ball's final tangential speed. Try solving for R, the length of the shortened rope.

7. Jan 22, 2007

### Koolaid

well w=v/r

so
if L_(initial) =L_(final)
and I=mr
and L=Iw

so

mrw=mrw
w=v/r
mrw=mr(v/r)
i canceled the m's out cause the mass is the same
i also canceled the r's out on the right side
r(1.26)=6
so from that we get
r=4.76
but were not trying to get r were trying to get a length?

mlw=ml(v/r)
i canceled the m's out cause the mass is the same
4(1.26)=l(6/4.76)

l=4 again.. which doesn't make since.. ugh i am so lost

8. Jan 22, 2007

### PhanthomJay

You said I =mr but of course you meant I =mr^2?
Then r_(initial) is 4, and r_(final) is what you are trying to solve for (where
r_(final) = length_(final). Give it another try....

9. Jan 23, 2007

### Koolaid

okay i got it 16(1.26)=r6 r=3.36

thanks a lot phantomJay for all your help