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, EP = mgl
Kinetic energy of the bob, EK = 0
Total energy = mgl … (i)
At the lowermost point (mean position): Potential energy of the bob, EP = 0
Kinetic energy of the bob, EK = (1/2)mv2
Total energy Ex = (1/2)mv2 ....(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.,
(1/2)mv2 = (95/100) mgl
∴ v = (2 × 95 × 1.5 × 9.8 / 100)1/2
= 5.28 m/s
On releasing the bob of pendulum from horizontal position, it falls vertically downward by a distance equal to length of pendulum i.e., h = l = 1.5 m .
As 5% of loss in P.E. is dissipated against air resistance, the balance 95% energy is transformed into K.E. Hence,
`1/2mv^2 = 95/100 xx mgh`
`=>v = sqrt(2xx95/100 xx gh) = sqrt((2xx95xx9.8xx15)/100) = 5.3 ms^(-1)`