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 a resistance?

Answer 1

Hey Friend,

◆ Answer - 5.34 m/s

◆ Explaination-
● Given-
l = 1.5 m
energy lost = 5%

● Solution-
a) In horizontal position-
P.E. = mgh = mgl
K.E. = 1/2 mv^2 = 0
Total energy E1 = mgl

But 5% energy is dissipitaed.
Thus E1' = 95% E1


b) At lowermost position-
P.E. = mgh = 0
K.E. = 1/2 mv^2
Total energy E2 = 1/2 mv^2

By law of conservation of energy
E1' = E2
95% mgl = 1/2 mv^2
v^2 = 95/100 × 10 × 1.5 × 2
v^2 = 28.5
v = 5.34 m/s

Thus, speed of the bob at lowermost position will be 5.34 m/s.

Hope that helps you...

Answer 2

Answer:

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