# Does time period of a simple pendulum depends upon the moment of inertia of the Bob of pendulum or its shape?

Time period of simple pendulum is given by

$$T=2\pi {\sqrt{\frac{L}{g}}}$$

Here L is the length of pendulum i.e. distance between the point of support to the centre of mass of the Bob.

Now consider a pendulum in which we replace spherical Bob with a disc having some mass. Rod is attached to the rim of the disc. The rod with which disc is attached is massless.

The time period of this pendulum should be

$$T=2\pi {\sqrt{\frac{L+r}{g}}}$$

because now distance between support or pendulum to the centre of mass of the disc is L+r.

But in the book they have calculated time period by considering moment of inertia of the disc and obtained a complex expression

$$T=2\pi {\sqrt{\frac{3r^2 +4rL+2L^2}{2g(r+L)}}}$$

Please help me understand why the expression for the time period of this pendulum with disc should not have same form as the time period for pendulum with spherical bob. We have only changed the the shape of the Bob, everything else is same.

There are approximations in $$T = 2\pi\sqrt{\frac{L}{g}}$$. It is what you get if the pendulum consists of a point mass on a massless string, and the amplitude is be small enough that $$sin(\Theta) \approx \Theta$$.

Here they are going one step beyond the usual approximation. The mass is no longer a point.

The mass rotates as the pendulum swings. This requires a torque. Part of the force of gravity goes into accelerating the mass and part into producing the torque.

Take the sum of the torques about the point A, you obtain

$$[I_D+m\,(r+L)^2]\,\ddot \theta+m\,g\,(r+L)\,\sin(\theta)=0\tag 1$$

where $$~I_D~$$ is the disc inertia about the center of the disc mass.

for a small angle $$~\sin(\theta)=\theta~$$ thus, equation (1)

$$\ddot\theta+\omega^2\,\theta=0$$

where $$\omega^2=\frac{m\,g\,(r+L)}{I_D+m\,(r+L)^2}$$

so if the disc inertia $$~I_D~$$ equal zero you obtain

$$\omega=\sqrt{\frac{g}{r+L}}\quad,T=\frac{2\pi}{\omega}$$

and with $$~I_D=m\,r^2~$$

$$\omega=\sqrt{\frac{g\,(L+r)}{2\,r^2+L^2+2\,L\,r}}$$

• You might check your calculation. You didn't add $I_D$ to the expression for $\omega^2$. Oct 3, 2023 at 13:56
• @mmeser $~I_D~$ appear at the denominator of $~ω^2~$ ?
– Eli
Oct 3, 2023 at 15:27
• I take it back. You did. But you are not getting the same expression as in the post. I see it. $I_D = 1/2mr^2$ Oct 3, 2023 at 18:03