A thin rod of length 082 m and mass 160 g is suspended freel

A thin rod of length 0.82 m and mass 160 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 5.74 rad/s. Neglecting friction and air resistance, find (a) the rod\'s kinetic energy at its lowest position and (b) how far above that position the center of mass rises?

Solution

a) KE =1/2*I*w^2 =1/2*(1/3)mr²*w² = ½(1/3)*.16*.82²*5.74² = 0.59077J

b) h=position of centre of mass: KE = mgh h = KE/mg = 0.59077/(.16*9.81) = .0.37638m