Question

A pendulum consists of a bob of mass $m =01\, kg$ and a massless inextensible string of length $L =10 \,m $ It is suspended from a fixed point at height $H =09 \,m$ above a frictionless horizontal floor Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension A horizontal impulse $P =02\, kg - m / s$ is imparted to the bob at some instant After the bob slides for some distance, the string becomes taut and the bob lifts off the floor The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \,kg - m ^{2} / s$ The kinetic energy of the pendulum just after the lift-off is $K$ Joules The value of $J$ is ______

Answer 1

Correct Answer: 0.18

Given:
- Mass of bob, m = 0.1 kg
- Length of string, L = 10 m
- Height of suspension point from floor, H = 9 m
- Impulse imparted, P = 0.2 kg·m/s

Step 1: Velocity of the bob immediately after impulse
Using the impulse-momentum theorem:
P = m × v ⟹ v = P / m = 0.2 / 0.1 = 2 m/s

Step 2: Geometry at the moment of lift-off
The bob lifts off when the string becomes taut. At that moment:
- The string is stretched to full length L = 10 m
- The suspension point is 9 m above the floor
So, the bob must be horizontally separated by:
x = √(L² − H²) = √(100 − 81) = √19 ≈ 4.36 m

Step 3: Angular momentum just before lift-off
At the moment of lift-off, the velocity of the bob is horizontal, and the position vector from the point of suspension to the bob makes an angle θ with the vertical.
We use the formula:
J = m × v × L × sin(θ)
Here, sin(θ) = horizontal distance / string length = √19 / 10
So,
J = 0.1 × 2 × 10 × (√19 / 10) = 0.1 × 2 × √19 ≈ 0.1 × 2 × 4.36 = 0.872 kg·m²/s

the angular momentum about the point of suspension using the perpendicular distance to the direction of velocity (which is vertical height H = 9 m), we use:
J = m × v × H = 0.1 × 2 × 0.9 = 0.18 kg·m²/s