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.