Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 6 s. Find d.

Answer 1

we know Time period of physical pendulum,

T = 2*pi*sqrt(I_support/(m*g*Lcm))

here I_support is the moment of inertia of the object about the pivot.

Lcm is the is the distance from pivot to center of mass of the object.

let m is the mass of the stick

so, I_support = Icm + m*d^2

= m*L^2/12 = m*d^2

= m*1^2/12 + m*d^2 (since L = 100 cm = 1 m)

= m/12 + m*d^2

Lcm = d

given T = 6 s

so,

T = 2*pi*sqrt( (m/12 + m*d^2)/(m*g*d) )

T = 2*pi*sqrt(1/12 + d^2)/(g*d))

T^2 = 4*pi^2*(1/12 + d^2)/(g*d)

T^2*g*d = 4*pi^2*(1/12 + d^2)

5^2*9.8*d = 4*pi^2*(1/12 + d^2)

on solving the above equation we get,

==> d = 0.0134 m or 13.4 cm <<<<<<<----------------Answer