A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L = 1.0 m . It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. I

Question

A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L = 1.0 m . It is suspended from a fixed point at height H = 0.9 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 = 0.2 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 liftoff is K Joules. The value of K is ____.

TestHub · Accepted Answer

The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is,

Angular momentum about top point $L = m V \times 0.9$

$L = 0.2 \times 0.9$

$\Rightarrow L = 0.18 kg m^{- 2} s$

Now after the bob slides for some distance, the string becomes taut and the bob lifts off the floor so when string get taut an impulse act by string on particle.

Due to impulse, component of $V$ along string get change and becomes zero but component of $V$ perpendicular to string remain same so,

$V_{1} = V \sin θ = 0$

Angular momentum will be conserved about the fixed point,

$L_{i} = L_{f} \Rightarrow 0.18 = 0.1 \times (V \cos θ) \times 1 \Rightarrow V \cos θ = 1.8 m s^{- 1} = V_{net (Final)}$

Now the kinetic energy just after lift up is given by,

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 0.1 \times 1 . 8^{2} \Rightarrow K E = 0.162 J$

Hence, $K = 0.162$

Physics - Rotation Question with Solution | TestHub

Doubts & Discussion