# Physical Pendulum , Torsional Pendulum

A physical pendulum is the generalized case of the simple pendulum. It consists of any rigid body that oscillates about a pivot point

Any rigid body suspended from a fixed support constitutes a physical pendulum . Consider the situation when the body is displaced through a small angle θ .

Torque on the body about O is given by

τ = mg l sinθ      ….. (i)

where l = distance between point of suspension and centre of mass of the body.

If I be the M.I. of the body about O.

Then τ = Iα      ….. (ii)

From (i) and (ii) , we get

$\displaystyle I \frac{d^2\theta}{dt^2} = -mgl sin\theta$

as θ and d2θ/dt2 are oppositely directed.

$\displaystyle \frac{d^2\theta}{dt^2} = -\frac{mgl}{I} \theta$

Since θ is very small.

Comparing with the equation $\displaystyle \frac{d^2\theta}{dt^2} = -\omega^2 \theta$ , we get

$\displaystyle \omega = \sqrt{\frac{mgl}{I}}$

$\displaystyle T = 2\pi \sqrt{\frac{I}{mgl}}$

### Torsional Pendulum

On rotating a body from its position of equilibrium, a restoring torque proportional to the angle of rotation comes into play, the body executes angular (or rotational) SHM .

Restoring Torque , $\large \tau = – C \theta$ ;Where C = Torsional Constant

If I = Moment of Inertia of body about vertical axis ,

$\large I \alpha = – C \theta$

$\large \alpha = – \frac{C}{I} \theta = – \omega^2 \theta$

$\large \omega = \sqrt{\frac{C}{I}}$

$\large T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{I}{C}}$

• A typical Torsinal pendulums is a disc suspended by a wire attached to the centre of mass of the disc. When the disc is rotated, the wire gets twisted and a restoring torque is produced in it. The disc, therefore, executes angular oscillations on being released.