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]