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kishin7

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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!

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!

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