Question

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of {{60}^{o}}
to the vertical, will be (If g=10m/{{s}^{2}}
) [MP PET 1996]
A) \frac{E}{4} B) \frac{3E}{4} C) \frac{\sqrt{3}}{4}E D) {{K}_{1}}

Answer 1

Apply energy conservation theorem,
energy at lowest position of Bob = energy , when Bob makes 60° to the vertical
1/2 mv² = 1/2 mv₁² + mgl(1 - cos60°)
Here v is speed at Lowest position , v₁ is speed , when it makes 60° with vertical and l is length of pendulum .
[Actually, height of Bob , when it makes 60° with vertical = l(1 - cos60°)]

∴ v² = v₁² + 2gl(1 - cos60°)
3² = v₁² + 2 × 10 × 0.5 (1 - 1/2)
9 = v₁² + 5
v₁² = 4 ⇒v₁ = 2m/s

Hence, speed of Bob = 2m/s

Answer 2

The speed of the Bob is 2m/s.