A particle falls freely near the surface of t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The angular momentum $L$ of the particle about point $O$ is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicu

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) The angular momentum $L$ of the particle about point $O$ is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicular distance from $O$ to the line of motion. Since the particle is falling freely,its speed $v$ increases with time,while $d$ remains constant. Therefore,the angular momentum $L$ increases.The moment of inertia $I$ of the particle about $O$ is $I = mr^2$,where $r$ is the distance from $O$ to the particle. As the particle falls,the distance $r$ decreases,so the moment of inertia $I$ decreases.The angular velocity $\omega$ of the particle about $O$ is given by $\omega = \frac{v_{\perp}}{r} = \frac{v \sin 	heta}{r}$. As the particle falls,$v$ increases,$\sin 	heta$ increases (since $	heta$ increases),and $r$ decreases. Consequently,the angular velocity $\omega$ increases.Since all statements are correct,the correct option is $D$.

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

Similar Questions

$A$ particle performs rotational motion with an angular momentum $L$. If the frequency of rotation is doubled and its kinetic energy becomes one-fourth,the new angular momentum becomes:

Choose the correct statement.

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of [where $g$ is gravitational acceleration]:

Vedclass Test Series

Exam Paper Generator

Online Exam Module