Angular acceleration problem

Punchlinegirl

A rod of length 57.0 cm and mass 1.90 kg is suspended by two strings which are 41.0 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.

I tried using torque= I* alpha
torque= L x f= (.57)(18.62)=10.6
I got from Newton's 2nd law, (9.8)(1.90)
so, 10.6 = I* alpha
I= (1/12)(mL^2)= (1/12)(1.90 * (.57)^2= .051
so, 10.6 = .051 alpha
alpha = 207.8 rad/s^2
alpha= a/L
207.8 = a / .57
a= 118.4 m/s ^2

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lightgrav

Homework Helper
First, the place that it is rotating around is at one end of the rod.
How far from this is the rod's center-of-mass?

Second,
Newton's 2nd law is "Sum of Forces = ma"
Maybe you mean Newton's 4th Law "Force by gravity = m g = m GM/r^2"

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