# Homework Help: Torque By a Mass on a Disk

1. Nov 19, 2007

### breez

1. The problem statement, all variables and given/known data
An 8.0-cm radius disk with a rotational inertia of 0.12 kg m2 is free to rotate on a horizontal axis. A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end. The mass is raised by using a crank to apply a 9.0-N m torque to the disk. The acceleration of the mass is:

2. Relevant equations
torque = I(ang. accel)

3. The attempt at a solution
I have no clue. I tried solving for the angular acceleration, but I don't know how to connect the acceleration of the mass to the acceleration of the disk.

2. Nov 19, 2007

### Astronuc

Staff Emeritus
See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/mi#cmi [Broken]

and

http://hyperphysics.phy-astr.gsu.edu/hbase/mi [Broken] - general formulae

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq - rotational quantities.

Given: 8.0-cm radius disk, so r = 0.08 m
rotational inertia of 0.12 kg-m2, so I = 0.12 kg-m2

Find the appropriate expression for I of a disk.

The disk is rotating on a horizontal axis. The suspended mass has a weight pulling it down, and the crank applies a torque to the disk and suspended weight at moment r.

What is the net moment on the suspended mass?

Last edited by a moderator: May 3, 2017
3. Nov 19, 2007

### breez

Is the moment arm of the force assumed to be r? From the question, I thought the string was attached to an arbitrary point on the disc, and a crank was spinning the mass on the other end of the string.

I don't know how to calculate the net moment. It seems that the net moment would depend on the length of the string.

Last edited: Nov 19, 2007
4. Nov 19, 2007

### breez

The linear acceleration of the mass should be the same as that of the disk. If we just find the linear acceleration of the disk with net torque 9 Nm, we get the answer.

I think the answer is 6 m/s^2

5. Nov 19, 2007

### Hells_Kitchen

The general idea how to solve these problems is the following:

Apply Newtons Second Law in Linear Motion --> Sum F = m*a
Apply Newtons Second Law for Rotational Motion --> Sum Torque = I * alpha

In any closed system there is the relationship that a = r*alpha, which means that the linear acceleration is the same for both the mass and the disk.

I believe that if you use those two steps it will be easier to understand whether your answer is correct or not.

I suspect that your answer is incorrect there should be torque from both the force that the gravity force of the mass(each in opposite directions)

6. Nov 19, 2007

### breez

Anyone have a possible solution? I feel this problem is impossible. The acceleration of the mass would undoubtedly be dependent on the length of string.

7. Nov 19, 2007

### Hells_Kitchen

Like i said above:
For the mass hanging on the string:
Sum F = m*a <=> T- mg = ma --> T = ma + mg

For the disk:
Sum Torque = I*alpha --> F*r - T*r = I * alpha or F*r - T*r = I*a/r

Substituting the value of the Tension we have above, we get:

F*r - (ma+mg)*r = I*a/r ==> a = (F*r - mgr)/((I/r) + m)

Now as much as i would like to plug in values i am usually satisfied with the variable solution only. I hope this helps!