# Angular acceleration and tension

1. Dec 3, 2007

### roldy

1. The problem statement, all variables and given/known data
A uniform rod of length L and mass m is supported as shown in the figure. A vertical cable at A suddenly breaks. Determine at this instant the angular acceleration of the rod and the tension in cables AE and BF.

2. Relevant equations
mass moment of intertia of bar: 1/12mL^2
summation of forces
relative accelerations

3. The attempt at a solution
attempt is in picture
stuck at this point
http://img228.imageshack.us/my.php?image=82wv8.jpg

2. Dec 3, 2007

### Gokul43201

Staff Emeritus
I don't see the "vertical cable at A" in the picture. Is the picture drawn after this cable has broken?

3. Dec 3, 2007

### roldy

Yes, vertical cable at A is not drawn in because it has broken. I don't think this cable works into the equations at all.

4. Dec 3, 2007

### Gokul43201

Staff Emeritus
No, you definitely need to use that information. The tensions in the other cables at the instant of breaking depends on the existence of the vertical cable at A. A fraction of a second later, the tensions in the other cables will readjust to make up for the loss of the first cable. You want to study the system before this readjustment happens.

First draw the picture with the vertical cable at A. You will immediately notice something (by inspection) about the tension in cable AE. You can also solve for the tension in BF. You then need to use these values and set the tension in the vertical cable at A = zero to determine the instantaneous acceleration of the rod.

5. Oct 11, 2011

### Jasonofcompsc

The tension in the rod doesn't matter because tension cannot effect angular acceleration. To solve this problem you basically have the mass move about the rod. Force of gravity translates into torque. Torque translates into angular acceleration. Technical you could do it without knowing the mass either.