Question

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5m$, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?

Answer 1

Length of the pendulum, $l=1.5m$

Mass of the bob $=m$

Energy dissipated $=5$%

According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position A:

Potential energy of the bob, $E_P=mgl$

Kinetic energy of the bob, $E_K=0$

Total energy $=mgl$ ------ (i)

At the lowermost point (mean position) B:

Potential energy of the bob, $E_P=0$

Kinetic energy of the bob, $E_K=\frac{1}{2}m{v}^2$

Total energy $E_x=\frac{1}{2}m{v}^2$ ....(ii)

As the bob moves from the horizontal position to the lowermost point, $5$% of its energy gets dissipated.

The total energy at the lowermost point is equal to $95$% of the total energy at the horizontal point, i.e.,

$\frac{1}{2}m{v}^2=\frac{95}{100}mgl$

$v={\left(\frac{2\times 95\times 1.5\times 9.8}{100}\right)}^{1/2}$

$v=5.28m/s$