# Torque, angular momentum, moment of inertia

## Homework Statement

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?

## Homework Equations

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

## 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.

Doc Al
Mentor
Wow I'm really lost with this one. Here's what I tried.
I=.5MR^2+2mr^2
What's the rotational inertia of a rod?

F=m$$\omega$$$$^{2}$$r
100=2(.4)$$\omega$$$$^{2}$$(.4)
$$\omega$$=17.667
Why multiply by 2? The tension acts on each mass separately.

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

The symbols appear to be in superscript but they should not be.
That's because you're mixing Latex and non-latex. To do that, use the in-line tags: itex instead of tex.

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.

Doc Al
Mentor
I=(1/12)MR^2+2mr^2
OK.

So do I use work = torque * theta to find work in this situation?
But you don't have enough information to find torque. (You can still do it that way, since the actual torque doesn't matter.)

F=ma
F=m * omega^2 * r
100=(.4) * (omega^2) * .4
OK. Note that F here is the tension in the string.

F=ma
F=m * R * alpha
100=(.4) * (.4) * alpha
Note sure what you're doing here. The tension is not directly related to the torque.

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

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.

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

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

I=(1/12)MR^2+2mr^2
I=.192 kg*m^2
Redo this calculation. (What's R?)

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?