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, 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, 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 = `1/2mv^2`

Total energy E_x = `1/2mv^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.,

K.E = `1/2mv^2` = `95/100` mgl

∴  v = `((2 × 95 × 1.5 × 9.8 )/ 100)^(1/2)`

= 5.28 m/s