**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 Î¸.

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 Î¸. 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)`**Torque**is defined as a twisting force that tends to cause rotation. Let, a body is facing a linear force F at an angle Î¸, at a distance r, measured from a fixed point (axis of rotation). This force will tend to cause rotation to the 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