# Deducing Expression of Potential Energy and Torque for a Simple Pendulum

Statement: With necessary figures deduce the following:
(a) Potential energy expression mgl (1-\cos\theta)
(b) Expression of torque mgl\sin\theta

Solution

In a simple pendulum with no friction, mechanical energy is conserved. Total mechanical energy is a combination of kinetic energy and gravitational potential energy. As the pendulum swings back and forth, there is a constant exchange between kinetic energy and gravitational potential energy.

(a)  For deducing, consider a simple pendulum. Let the bob of mass m be released after being given an angular displacement θ.

Let, Length of the pendulum, OA = OB = l.

From the figure we can see, the pendulum has been shifted from point A to B at an angular displacement θ. So, maximum potential energy will be stored in the bob on point B if the velocity reaches zero and kinetic energy is zero at that point.

Though the mass has been shifted from point A to point B, it has been lifted at a height AC = h with respect to the datum position A.

We know,

Potential Energy = mass \times acceleration due to gravity \times height

\therefore E_p=mgh .......................................... (1)

From figure, \triangle OCB is a right angled triangle.

Hence,

\cos\theta=\frac{OC}{OB}=\frac{OC}l

\therefore OC=l\cos\theta ..................................(2)

Now,

OC = OA – AC

l\cos\theta=l-h    [From (2)]

h=l-l\cos\theta

h=l(1-\cos\theta) ................................................(3)

From equation (1) we can write,

E_p=mgl(1-\cos\theta)

Hence, Potential energy expression mgl (1-\cos\theta)
[Deduced]

(b) Torque is defined as a twisting force that tends to cause rotation. Let, a body is facing a linear force F at an angle θ, at a distance r, measured from a fixed point (axis of rotation). This force will tend to cause rotation to the body.
Fig - 2: Linear Force applied to a body

Torque is defined as,

\vec \tau =\vec r \times \vec F ........................... (4)

The cross product is defined to satisfy the following property,

|\vec a\times\vec b|=ab\sin\theta  .....................(5)

Therefore,

|\vec \tau |= |\vec r \times \vec F|=Fr\sin\theta ...................... (6)

Where, \tau = Torque

F = Linear force

r = distance measured from the axis of rotation to where the linear force is applied

Now, in Figure – 1, the force that acts on the bob is gravitational force mg. Due to this force the bob starts swinging to the left from position B. The pendulum rotates about point O and the force is applied at a distance l , from the axis of rotation.

Hence, F=mg \sin\theta and r = l

Therefore, from equation (6),
Torque, \tau = Fr \sin\theta
\tau = mgl\sin\theta

So, the expression of torque is mgl\sin\theta.
[Deduced]