# Can anyone solve this?

1. Feb 4, 2005

### sebasalekhine7

A uniform rod has a length of 2m. It is hinged to a wall at the left end and held horizontally by a vertical massless string at the right end. What is the angular acceleration of the rod the moment the string is released?

2. Feb 4, 2005

### rpc

angular accel = tangential accel (in this case gravity) / radius

3. Feb 4, 2005

### Curious3141

Easiest way : $$\tau = I\alpha$$, which means torque = moment of inertia*angular acceleration.

What is the moment of inertia of a rod of uniform length about the end ?

What is the torque exerted by the weight of the rod about the end ?

4. Feb 4, 2005

### sebasalekhine7

I think that $$I=1/3(ML^2)$$

5. Feb 4, 2005

### Curious3141

Correct. You don't have to know the derivation, but it's fairly simple using a little calculus.

What's the torque about the end ?

6. Feb 4, 2005

### sebasalekhine7

Ok I got it, thanks, $$\tau = I\alpha$$ when $$\tau =1/2mgL$$ and therefore $$\frac{1}{3}(ML^2)\alpha=1/2mgL$$ and angular acceleration ends up being 7.35.

You see, the problems I post here are the ones I am studying for the next physics science league, is a competition between schools here in NJ.

Last edited: Feb 4, 2005
7. Feb 4, 2005

### Curious3141

Excellent, you got it (except both your m's need to be the same case) !

8. Feb 5, 2005