# Torque, angular momentum, moment of inertia

1. Apr 28, 2010

### Johnny0290

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

Consider a hollow tube of mass M = 1.2 kg and length L = 1.6 m that rotates about an axle through its center and perpendicular to its length. Inside the tube are two masses, m_1 = 0.4 kg each. These masses are initially held a distance d = 0.8 m apart by a string and centered in tube. The maximum tension the string can sustain is 100 N. You may consider that the radius of the tube is negligible (i.e. its moment of inertia is that of a 'stick') and that the masses held by the string are point-like.

1.1 Starting from rest, the cylinder starts to rotate as a result of a constant driving torque applied to it. What is the work done by this torque up to the point at which the string breaks?

2. Relevant equations

Work=$$\tau$$$$\theta$$
$$\tau$$=rxF=I$$\alpha$$
F=ma=m$$\frac{v^{2}}{r}$$=m$$\omega$$$$^{2}$$r

3. The attempt at a solution

Wow I'm really lost with this one. Here's what I tried.
I=.5MR^2+2mr^2

F=m$$\omega$$$$^{2}$$r
100=2(.4)$$\omega$$$$^{2}$$(.4)
$$\omega$$=17.667

F=m$$\alpha$$R
100=2(.4)$$\alpha$$(.4)
$$\alpha$$=312.5

Definitely feel like I'm doing something wrong here.^^^

So with that I could find the torque using torque=I*alpha, but I would still need to somehow find the angle theta. I'm pretty sure I'm approaching this problem wrong.

Any help is appreciated!! Thank you. Sorry about the formatting, I'm not sure how to put the symbols in correctly.
The symbols appear to be in superscript but they should not be.

2. Apr 28, 2010

### Staff: Mentor

What's the rotational inertia of a rod?

Why multiply by 2? The tension acts on each mass separately.

You don't need the torque, just the work done.

That's because you're mixing Latex and non-latex. To do that, use the in-line tags: itex instead of tex.

3. Apr 28, 2010

### Johnny0290

I=(1/12)MR^2+2mr^2

So do I use work = torque * theta to find work in this situation?

F=ma
F=m * omega^2 * r
100=(.4) * (omega^2) * .4

F=ma
F=m * R * alpha
100=(.4) * (.4) * alpha

omega^2 = omega_0^2 = 2*alpha*theta

W=torque*theta
W=I*alpha*theta
W=60Joules

I'm not really sure how I could find the work any other way.

4. Apr 28, 2010

### Staff: Mentor

OK.

But you don't have enough information to find torque. (You can still do it that way, since the actual torque doesn't matter.)

OK. Note that F here is the tension in the string.

Note sure what you're doing here. The tension is not directly related to the torque.

Hint: When work is done, what happens to the energy?

5. Apr 28, 2010

### Johnny0290

Okay so I could say that the change in kinetic energy is equal to the work done on the rod by the torque.

KE_rotational=.5 * I * omega^2

F=ma
T=m * omega^2 * r
100=(.4) * (omega^2) * .4

Not sure if I found omega correctly. This is the max radial velocity before the string breaks?

I=(1/12)MR^2+2mr^2
I=.192 kg*m^2

KE=W=(.5)*(.192)*(25^2)
W=60

Something tells me I'm still doing something wrong since I got the same answer.

6. Apr 28, 2010

### Staff: Mentor

Exactly.

It's the max angular speed before the string breaks.

Redo this calculation. (What's R?)

7. Apr 28, 2010

### Johnny0290

Oh it was supposed to be L not R.

I=[(1/12) * (1.2) * (1.6^2)] + [(2) * (.4) * (.4^2)]
I=.384

KE=W=(.5)*(.384)*(25^2)
W=120

Does this look good?

8. Apr 29, 2010

### Staff: Mentor

Looks good now.

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