Moment of Inertia/Acceleration of a Swinging Rod

[kishin7]

I've been at this problem for almost an hour and I know that I'm going in the right direction but I think that I'm using something wrong within the calculations and what not. But well...here's the problem:

A rod of length 61.0 cm and mass 1.10 kg is suspended by two strings which are 44.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

http://homework.inst.physics.ucsb.edu/msuphysicslib/Graphics/Gtype20/prob13a_1016full.gif

What I've done so far is that I found the downward FORCE for the rod which is (1.10 kg)(9.81 m/s^2) = 10.80N. I then assumed that once the end of B is cut the TORQUE would then be T=L x F ==> T= 0.61m * 10.80N = 6.59 Nm. The moment of Inertia of this rod is I= (1/3)ML^2 and that equals 0.136 kgm^2. I then plugged the MOMENT OF INERTIA and TORQUE into the equation T = I x a(alpha) and solved for the angular acceleration which I got (alpha) = 6.59/0.136 = 48.46 rad/s^2. Since I know alpha = a/r I used the R value of .44m and tried to solve for acceleration but it is wrong.

Help is much appreciated!

 

[Galileo]

You can consider the gravity to act in the COM of the rod, which is the middle, so the torque is not LxF, but 1/2LxF.

 

[thepatientmental]

First of all, great job on your approach to the problem so far! Your understanding of the concept of moment of inertia and torque is correct.

However, there is one mistake in your calculation. When calculating the torque, you used the length of the rod (0.61m) instead of the length of the string (0.44m). This is because the torque is being exerted on the rod by the string, not the rod itself.

So the correct torque would be 0.44m * 10.80N = 4.75 Nm.

Using this value in your equation T = I * alpha, we get:

4.75 Nm = (0.136 kgm^2) * alpha

Solving for alpha, we get 34.93 rad/s^2.

Now, to find the acceleration of end B, we can use the formula a = r * alpha. Plugging in the values, we get:

a = (0.44m) * (34.93 rad/s^2) = 15.37 m/s^2

So the magnitude of the initial acceleration of end B is 15.37 m/s^2.

I hope this helps! Keep up the good work.