Question

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, 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, $\text I$ = $1.5 \,m$
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:
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): Potential energy of the bob, $E_p$ = $0$
Kinetic energy of the bob, $E_k$= $\frac{1}{2}mv^2$
Total energy $E_x$= $\frac{1}{2}mv^2$ … (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.,

$\frac{1}{2}$ $mv^2$ = $\frac{95}{100}\times\,mgl$

$\therefore$ $v$ = $\sqrt{\frac{2\times 95\times 1.5\times 9.8}{100}}$=

$5.28\;m/s$