# Angular acceleration & cylinder

1. Nov 6, 2005

### Punchlinegirl

M, a solid cylinder (M=1.59 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.750 kg mass, i.e., F = 7.357 N. Calculate the angular acceleration of the cylinder.
I used,
I*alpha= mgr
(1/2)mr^2 *alpha= mgr
(1/2)(1.59)(.127)^2 * alpha= (1.59)(9.8)(.127)
Solving for alpha gave me 155.9 rad/s^2
which wasn't right. Can someone help? Thanks.

2. Nov 6, 2005

### lightgrav

Did you forget the "rotational Inertia" of the hanging mass?
Include the .75 kg mass at the R^2 where the string joins.

Otherwise, you have to use the Tension in the string, rather than mg,
to provide torque on the disk. (with mg - T causing ma of the hanger).

3. Nov 6, 2005

### Punchlinegirl

so would I do..
(1/2)(1.59+ .75)(.127)^2 * alpha= (1.59)(9.8)(.127)

I'm a little confused about where the .75 comes into it

4. Nov 6, 2005

### lightgrav

IF you actually HAVE a hanging mass on the string you'd do
I_total = I_disk + I_hanger = (1/2)(1.59)(.127)^2 + (.75)(.127)^2 .

But the wording in the problem is peculiar, you might NOT have a hanger.

You somehow used the weight of the cylinder (rather than 7.36 N) to provide the torque which is supposed to angularly accelerate the cylinder.
Sorry I hadn't noticed it before, the wording is really distracting.

5. Nov 7, 2005

### Punchlinegirl

Ok I figured out what I was doing wrong and got the right answer.. it was 72.9 rad/s^2.
The second part says if instead of the force F an actual mass m= .750 kg is hung from the string, find the angular acceleration of the cylinder. I got this part it was 37.5 rad/s^2.
The third question says how far does m travel downward between 0.530 s and 0.730 s after the motion begins?

I used a= delta w/delta t
37.5= delta w/ .2