
#1
Nov605, 08:47 PM

P: 225

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 This isn't right... can someone please help me? 


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