# Find the angular acceleration of a hinged beam

1. May 13, 2013

Hi,
I have the following problem that I'm lost on.

The answer is 41.4 rad/s2 but I don't understand how to arrive at it.

2. Relevant equations
ƩT=I(alpha)
I=1/3 ML2 for a rod with axis through one end.

3. The attempt at a solution
I solved the above equations to get (alpha)=T/I, but no matter what I try from here I get small numbers under 5.
I'm confused on a couple points: is the center of gravity in the middle of the rod or the tip? Should I add the mass of the box (since the cable is cut, not the rope)? And I'm really confused as to how to write the sum of torques.
I tried following this example, with and without the box, but it I'm getting small numbers not even close to 41. Could someone please explain clearly to go about solving this? Thank you!

Last edited: May 13, 2013
2. May 13, 2013

### haruspex

3. May 13, 2013

I converted the masses to newtons to get 833N for the beam and 4214N for the box. Then I used trig to get the length of the beam (3m) and the angles: 37degrees below the beam and 53 above. Then I tried this, based on an example from my class notes (assuming the center of the beam's gravity is in the middle):
,
substituting 5047N for m and 3 for L, then solving for alpha. But that's clearly wrong.
Then I tried following this example exactly as shown, with units as shown, but I got 4.4(without the box) and 3.9 (with the box's weight added). Not even close to 41.

Last edited: May 13, 2013
4. May 13, 2013

### haruspex

You're overlooking that the mass will accelerate too. Let the tension in the rope be T. Develop the free body equations for the beam and the mass separately. The relationship between the beam's angular acceleration and the mass's vertical acceleration will need a bit of care.
I recommend doing all working symbolically, only plugging in numbers as the final step. That goes for the angles too. It makes it much easier to follow the reasoning and to spot mistakes.