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

Given, the length of the pendulum is 1.5  m and energy dissipated against air resistance is 5% of the initial energy.

Let $U_h$ be the potential energy of the bob, $K_h$ be the kinetic energy of the bob, $L$ be the length of the pendulum, $m$ be the mass of the bob and $g$ be the acceleration due to gravity.

The energies of the pendulum at horizontal position are $U_h=mgl$ and $K=0$.

Therefore, total energy $T_h$ in the case of horizontal position is,

$T_h=U_h+K_h=mgl+0=mgl$ …… (1)

Let $U_l$ be the potential energy of the bob and $K_l$ be the kinetic energy of the bob at the lowermost point, then

$U_l=0$

$K_l=\frac{1}{2}m{v}^2$

Here, $v$ is the final speed of bob at lowermost point.

Therefore, total energy $T_l$ in the case of lowermost position is,

$T_l=U_l+K_l=0+\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^2$ …… (2)

It is given that 5% of the initial energy gets dissipated due to air resistance; hence the energy at lowermost point is only 95% of the initial value. Thus,

$T_l=\frac{95}{100}\times T_h$

From equation (1) and (2),

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

Substitute the values in the above expression.

$\frac{1}{2}m{v}^2=\frac{95}{100}\times m\times 9.8\times 1.5$

$v=2\times 9.8\times 1.5\times 95100\approx 5.3m/s$

Hence, the final speed of bob at lowermost point is 5.3  m/s .