Question

A small mass m is attached to a massless string whose other end is fixed at P as shown in figure. The mass is undergoing circular motion in x-y plane with centre O and constant angular speed $\omega$. If the angular momentum of the system, calculated about O and P and denoted by ${\overrightarrow{L}}_{O}$ and ${\overrightarrow{L}}_{P}$ respectively, then

• A. ${\overrightarrow{L}}_{o}$ and${\overrightarrow{L}}_{p}$ do not vary with time.
• B. ${\overrightarrow{L}}_{o}$ varies with time while${\overrightarrow{L}}_{p}$ remains constant.
• C. ${\overrightarrow{L}}_{o}$ remains constant while ${\overrightarrow{L}}_{p}$ varies with time.
• D. ${\overrightarrow{L}}_{o}$ and ${\overrightarrow{L}}_{p}$both vary with time.

Answer 1

Answer: (C)

${\overrightarrow{L}}_{o}$ remains constant while ${\overrightarrow{L}}_{p}$ varies with time.

Answer 2

In figure, weight mg of the mass acts vertically downwards and tension T in the string acts along KP. Two rectangular components of T are $T\cos\theta$opposite to mg and $T\sin\theta$along KO.

About O, torques due to T $\cos\theta$and mg cancel out. Torque due to $T\sin\theta$is zero. Therefore, net torque about O is zero.

As ${\overset{\rightarrow}{\tau}}_{P} = \frac{d{\overset{\rightarrow}{L}}_{P}}{dt} \neq 0.$

$\therefore$ ${\overset{\rightarrow}{L}}_{P}$is not central force is zero