Question
Question: The time period of a bar pendulum when suspended at distances 30 cm and 50 cm from its centre of gra...
The time period of a bar pendulum when suspended at distances 30 cm and 50 cm from its centre of gravity comes out to be the same. If the mass of the body is 2kg. Find out its moment of inertia about an axis passing through the first point.
A) 0.24 kg m2
B) 0.72 kg m2
C) 0.48 kg m2
D) data insufficient
Solution
We are given a bar pendulum and when it is suspended at two lengths, the centre of gravity comes out to be the same. Given the mass is 2 kg. Angular oscillations of rigid bodies in the form of a long bar can be used to find acceleration due to gravity.
Complete step by step answer:
The bar pendulum is free to oscillate about the knife-edge as an axis. The restoring torque τ for an angular displacement θ is given by,
τ=mglsinθ
Since the angle is very small, it can be written as τ=mglθ
Substituting the values,
For first case: τ1=mg×0.3×θ
For second case: τ2=mg×0.5×θ
Let the time period of the pendulum be T, by using the parallel axis theorem,
Moment of inertia is
I1=I+2×0.32 ⟹I2=I+2×0.52
Because time period of blue points is equal,
I1=I+2×0.32 ⟹I2=I+2×0.52 ⟹2πτ1I1=2πτ2I2 ⟹τ1I1=τ2I2
Putting out the values of the moment of inertia,
I+2×0.52I+2×0.32=0.5mg0.3mg ⟹I+2×0.52I+2×0.32=53 ⟹I=0.30kgm2
So, the value of moment of inertia comes out to be 0.30kgm2
moment of inertia about an axis passing through first point can be find as
I1=0.3+2×0.32 =0.48kgm2
So, the correct answer is “Option C”.
Note:
An ideal simple pendulum cannot be realized under laboratory conditions. Hence, a compound pendulum is used to determine the acceleration due to gravity in the laboratory. It is widely used and an accurate experiment as direct measurement of the acceleration due to gravity is very difficult.