Consider a thin 24 m rod pivoted at one end. A uniform density spherical object

Question

Consider a thin 24 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 4.9 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I(rod) =(1/3)m L^2 and the moment of inertia of the sphere about its center of mass is I(sphere) =(2/5)m r^2. What is the angular acceleration of the rod immediately after it is released from its initial position of 41 degrees from the vertical? The acceleration of gravity g = 9.8 m/s^2. Answer in units of rad/s^2

Explanation / Answer

It seems the rod has also mass of 9 kg. It has center of mass at 7m from the pivot. So there are two mass points, one, the rod of 9 kg at 7m from the pivot and the other the the sphere of 9 kg at a distance 24m + 4.9m =28.9m from the pivot. we have to find out the CM of the two body system. if it is at x m from the pivot, then (x - 7)m . 9kg = (28.9 - x) . 9kg . Hence x = 17.95m. The entire 9kg +9kg = 18 kg mass may be supposed to be concentrated at this point, 17.95m from the pivot. So the moment of inertia of the system having mass 18kg at a distance 17.95 m from the pivot is I = MR^2 = 18 kg. (17.95).(17.95) m^2 . Force on the system is the weight 18kg.9.8m/s^2 acting downwards. So the torque is L = (17.95m). 18kg.9.8m/s^2 sin 35^0 = I . angular acceleration. Now you do the rest.

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