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?
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"