A solid cylinder of mass 2 kg and radius 0.2 | vedclass

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

Question

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular v

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.For a solid cylinder rotating about its own axis,the moment of inertia is $I = \frac{1}{2}MR^2$.Given: Mass $M = 2\ kg$,Radius $R = 0.2\ m$,Angular velocity $\omega = 3\ rad/s$.Substituting the values:$I = \frac{1}{2} 	imes 2\ kg 	imes (0.2\ m)^2 = 1 	imes 0.04 = 0.04\ kg\cdot m^2$.Now,calculate the angular momentum:$L = I\omega = 0.04\ kg\cdot m^2 	imes 3\ rad/s = 0.12\ J-s$.Therefore,the angular momentum of the cylinder before collision is $0.12\ J-s$.

Similar Questions

$A$ particle is rotating in a circle with uniform speed as shown. The angular momentum of the particle with respect to the origin is .........

Angular momentum is

Write the general formula of total angular momentum of rotational motion about a fixed axis.

$A$ fan of moment of inertia $0.6\,kg \cdot m^2$ is turned up to a working speed of $0.5$ revolutions per second. The angular momentum of the fan is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module