A small ball of mass M is attached to the end of a uniform rod of equal mass M a

Question

A small ball of mass M is attached to the end of a uniform rod of equal mass M and length L = 2.0 m that is pivoted at the top. Calculate the period of oscillation for small displacements from equilibrium. Hint: Moment of inertia of this physical pendulum is I = I rod + I ball = 1/3 ML2 + ML2. When it is displaced from the equilibrium position, the torque caused by its weight is tau = tau rod + tau ball = - Mg middot 1/2 L middot sin theta - Mg middot L middot sin theta For small angles, sin theta = theta, tau = tau rod + tau ball = -Mg middot 1/2 L middot theta - Mg middot L middot theta tau = I middot alpha = I middot theta..

Explanation / Answer

for a pendulum

w^2 = mgd /I

In your case

I = 1/3 M L^2 + ML^2 = (4/3) M L^2

And

d is the distance from the pivot to the center of mass. The center of mass is the avg position of the center of the rod and the center of the ball, since they have equal mass. So d is 3/4 the length of the rod, or d = 1.50 m

and

w^2 = M g d / (4/3) M L^2 = 3 g d / 4 L^2 = 3*9.8*1.5/4*2^2 = 2.75625

so w = 1.66012

and the period is

T = 2pi / w = 2pi / 1.66012 = 3.785 seconds

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