A thin rod of length 0.83 m and mass 110 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 5.71 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.

Explanation:

(a)  The given data is as follows.

    Length of the rod, L = 0.83 m

    Mass of the rod, m = 110 g = 0.11   (as 1 kg = 1000 g)

 At the lowest point, angular speed of the rod () = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

      I =

        =

        =

        = 0.00631 + 0.415

        = 0.42131

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

             K =

                =

                = 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b)   According to the conservation of total mechanical energy of the rod, we have

or,      mgh = K = 6.86 J

Therefore,      h =

                          =

                          = 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

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