A uniform rod of length 1.29 m is attached to a frictionless pivot at one end. It is released from rest from an angle q = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass. [tex] I= (1/2)ML^2 [/tex] I= (1/2)(1.29^2), since mass isn't given I just used 1 kg. I=.832 kg*m^2 [tex] \tau= rFsin \theta [/tex] =.645(9.8)(sin 21)= 2.27 Nm [tex] \tau = I \alpha [/tex] 2.27= .832 [tex] \alpha [/tex] [tex] \alpha [/tex] = 2.73 rad/s^2 [tex] a= \alpha * r [/tex] a= 2.73 * .645 a= 1.76 m/s^2 This isn't right.. can someone help me?