The moment of inertia of a solid cylinder abo | vedclass

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

Question

(D) For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.Since the c

Vedclass · Accepted Answer

$\frac{3}{2}I\omega^2$

Vedclass · Answer

(D) For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.Since the cylinder rolls without slipping,the linear velocity $v$ is related to the angular velocity $\omega$ by $v = R\omega$.The total kinetic energy $(K.E.)$ of a rolling body is the sum of its translational kinetic energy and rotational kinetic energy:$K.E. = K.E._{trans} + K.E._{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.Substituting $v = R\omega$:$K.E. = \frac{1}{2}M(R\omega)^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2\omega^2 + \frac{1}{2}I\omega^2$.Since $I = \frac{1}{2}MR^2$,we have $MR^2 = 2I$.Substituting this into the equation:$K.E. = \frac{1}{2}(2I)\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2 + \frac{1}{2}I\omega^2 = \frac{3}{2}I\omega^2$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

Similar Questions

$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

$A$ plane is inclined at an angle of $30^{\circ}$ with the horizontal. If a sphere rolls down this plane without slipping,what will be the linear acceleration of the sphere?

$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

$A$ ring takes time $t_1$ in slipping down an inclined plane of length $L$ and takes time $t_2$ in rolling down the same plane. The ratio $\frac{t_1}{t_2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module