Question

A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 3.0 m/s when at the release point, to what maximum angle will it move in the second half of its swing?

a. 47°

b. 36°

c. 30°

d. 19°

Answer 1

By using energy conservation bet ween release point and at maximum angle,

KEi + PEi = KEf + PEf

here, KEi = kinetic energy at release point = 0.5*m*v^2

PEi = Potential energy at release point = m*g*hi

KEf = final kinetic energy = 0

PEf = maximum potential energy = m*g*hf

given, v = initial speed = 3.0 m/s

hi = initial height = L(1 - cos)

L = length of pendulum = 2.0 m

= initial angle = 25 deg

Let, maximum angle = max

then, hf = final height = L*(1 - cos(max))

So,

0.5*m*3^2 + m*g*2*(1 - cos 25 deg) = 0 + m*g*2*(1 - cos(max))

1 - cos(max) = 0.5*3^2/(2*g) + 1 - cos 25 deg

1 - (0.5*3^2/(2*g) + 1 - cos 25 deg) = cos(max)

cos(max) = - 0.5*3^2/(2*9.81) + cos 25 deg

max = arccos(0.67695)

max = 47 deg

So, correct option is A.