Oscillations L10 Physical Compound Pendulum, Torsional Pendulum & Intro to Rolling SHM

In Oscillations Lecture 10, we give a compact introduction to three important angular SHM systems:

Physical (compound) pendulum – any rigid body oscillating about a horizontal axis.
Torsional pendulum – body suspended by a twisting wire.
Rolling SHM – a solid cylinder attached to a spring and rolling without slipping.
The aim is to show how the same SHM pattern appears again and again:

Restoring torque proportional to angle,
Angular acceleration alpha = minus omega squared theta,
Time period T = 2 pi by omega.
1. Physical (Compound) Pendulum
Definition: Any rigid body suspended from a fixed horizontal axis and allowed to oscillate in a vertical plane.

Distance from pivot to centre of mass = L (small L)
Mass = m
Moment of inertia about pivot = I (capital I)
For small angular displacement theta:

Gravitational torque:
tau = minus m g L sin theta approximately equal to minus m g L times theta.

Rotational equation:
I d2theta by dt2 = tau.

So:

d2theta by dt2 = minus (m g L divided by I) theta
Hence
omega = square root of (m g L divided by I)
T = 2 pi square root of (I divided by m g L)
Using the parallel axis theorem:

I = I cm plus m L squared, with I cm = m k squared (k = radius of gyration about CM)
We get:

T = 2 pi square root of (k squared plus L squared divided by L g)
Define equivalent length of a simple pendulum:

L eq = k squared divided by L plus L
So T = 2 pi square root of (L eq divided by g)
Minimum period occurs when L equals k:

T min = 2 pi square root of (2 k divided by g)
Example (metre stick):

Uniform rod length 1 m, pivot at 75 cm mark.
Distance CM to pivot d = 0.25 m
I cm = m L rod squared by 12
I about pivot = I cm + m d squared
Use T = 2 pi square root of (I divided by m g d).
2. Torsional Pendulum
Rigid body suspended by a wire. When twisted by angle theta:

Restoring torque:
tau = minus k theta (k = torsion constant of wire)
Rotational equation:

I d2theta by dt2 = minus k theta
So d2theta by dt2 = minus (k divided by I) theta
Hence:

omega = square root of (k divided by I)
T = 2 pi square root of (I divided by k)
No small angle restriction is needed as long as the wire remains within its elastic limit.

Example (disc on wire):

Uniform disc: m = 0.200 kg, r = 0.050 m
I = m r squared by 2 = 2.5 times 10 power minus 4 kg m squared
Given T = 0.20 s
From T = 2 pi square root of (I divided by k):
k = 4 pi squared I by T squared approximately 0.25 kg m squared per s squared.
3. Rolling SHM: Solid Cylinder Attached to a Spring
Solid cylinder: mass m, radius R
Spring constant k, horizontal surface, rolling without slipping
Small rotation theta:

Centre displacement x = R theta
Spring extension x = R theta
Spring force F s = k x = k R theta (towards mean)
Equations used (not fully derived in detail):

Translation: F s minus friction f = m a
Rotation about CM: f R = I cm alpha
Rolling without slipping: a = R alpha
For solid cylinder: I cm = (1 by 2) m R squared
From these:

alpha = minus (2 k divided by 3 m) theta
So omega = square root of (2 k divided by 3 m)
Time period:
T = 2 pi square root of (3 m divided by 2 k)
This shows how combined translation plus rotation can still give simple harmonic motion.

Key Takeaways
Physical pendulum:
T = 2 pi square root of (I divided by m g L),
with I = I cm + m L squared and equivalent length L eq = k squared divided by L plus L.

Torsional pendulum:
tau = minus k theta,
T = 2 pi square root of (I divided by k).

Rolling SHM (solid cylinder plus spring, pure rolling):
T = 2 pi square root of (3 m divided by 2 k).

In all these cases, the SHM form comes from:

Restoring torque proportional to angular displacement,
Angular SHM equation d2theta by dt2 plus omega squared theta equals zero.

Видео Oscillations L10 Physical Compound Pendulum, Torsional Pendulum & Intro to Rolling SHM канала CBSE & JEE Physics | Dr Kedar Pathak