Question

A meter stick that has a mass of 90 grams is suspended from the 20 centimeter mark and set swinging freely.

1.Calculate the rotational inertia.

2. Calculate the period of the resulting small amplitude oscillation.

Answer 1

given
m = 90 g = 0.09 kg
L = 1 m
distance between center of mass and axis of rotation, d = 0.5 - 0.2

= 0.3 m
1) The rotational inertia, I = Icm + m*d^2 (using parallel axis theorem)

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

= 0.09*1^2/12 + 0.09*0.3^2

= 0.0156 kg.m^2 <<<<<<<<<<--------------------Answer

2) Time period of the physical pendulum,

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

= 2*pi*sqrt(0.0156/(0.09*9.8*0.3))

= 1.52 s <<<<<<<<<<--------------------Answer

Note :

According to Parallel axis theorem moment of inertia of the rod about given axis,

I = Icm + m*d^2

here d is the distance between axis of rotation to center of mass = 50 cm - 20 cm

= 30 cm

= 0.30 m

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

= 0.09*1^2/12 + 0.09*0.3^2

= 0.0156 kg.m^2