Question

A pendulum consists of a 320-g solid ball 15.0 cmin diameter, suspended by an essentially massless string 85.0 cm long.

Calculate the period of this pendulum, treating as a physical pendulum.

Answer 1

Time period of a physical pendulum is given by:

T = 2*pi*sqrt (I/(m*g*d))

I = moment of inertia of solid ball about rotation axis

Using parallel axis theorem:

I = I0 + m*L^2

I0 = moment of inertia of solid ball about it's center of mass = (2/5)*M*R^2

R = radius of solid ball = 15.0 cm/2 = 7.5 cm = 0.075 m

d = length of pendulum = L + R = 85.0 cm + 7.5 cm = 92.5 cm = 0.925 m (Remeber that we cannot ignore the increase in length due to radius of ball)

So,

T = 2*pi*sqrt [(2*m*R^2/5 + m*L^2)/(m*g*L)]

T = 2*pi*sqrt [(2*R^2/5 + L^2)/(g*L)]

Using given values:

T = 2*pi*sqrt [(2*0.075^2/5 + 0.925^2)/(9.81*0.925)]

T = 1.932 sec = time period of pendulum

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