A thin rod of length 1.2 m and mass 220 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 2.33 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.

Ki = 0.28665 J

h = 0.133 m

Explanation:

Given:

- The mass of rod M = 0.22 kg

- The length of rod L = 1.2 m

- The angular speed at the lowest point w = 2.33 rad /s

Find:

(a) the rod's kinetic energy at its lowest position

(b) how far above that position the center of mass rises.

Solution:

- The moment of inertia of a rod pivoted at one of its ends is given by I:

                                   I = ML^2 / 3

- The Kinetic energy at the lowest point is given by the rotational energy as follows:

                                   Ki = 0.5*I*w^2

                                   Ki = 0.5*ML^2 / 3*w^2

                                   Ki = ML^2*w^2 / 6

                                   Ki = (.22*1.2^2*2.33^2) / 6

                                   Ki = 0.28665 J

- Since no external force was acting on the rod we can apply the conservation of energy of system consisting of the rod where the change in kinetic energy leads to a change in gravitational potential Energy:

                                   Kf - Ki = Pf - Pi

                                   0 - 0.28665 J = mg( 0 - h )

                                   mg*h = 0.28665

                                   h = 0.28665 / ( 0.22*9.81 )

                                   h = 0.133 m

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