Ncert Exercises & Solutions

Question 6.18:

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?

Length of the pendulum, l = 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, $U_{p}$ = mgl

The kinetic energy of the bob, $E_{k}$ = 0

Total energy = mgl … (i)

At the lowermost point (mean position):

The potential energy of the bob, $U_{p}'$ = 0

The kinetic energy of the bob, $E_{k} = \frac{1}{2} {mv}^{2}$

Total energy $E_{T} = \frac{1}{2} {mv}^{2}$ … (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.,

$\frac{1}{2} {mv}^{2} = \frac{95}{100} \times mgl$
$\Rightarrow v = \sqrt{\frac{2 \times 55 \times 1.5 \times 9.8}{100}}$

= 5.28 m/s

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