A solid cylinder of mass M and radius R rolls | vedclass

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

Question

(C) By the law of conservation of energy,the potential energy at the top is equal to the total kinetic energy at the bottom.$PE = Mgh$$KE = \frac{1}{2

Vedclass · Accepted Answer

$\frac{2}{R} \sqrt{\frac{g h}{3}}$

Vedclass · Answer

(C) By the law of conservation of energy,the potential energy at the top is equal to the total kinetic energy at the bottom.$PE = Mgh$$KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.Substituting these into the kinetic energy equation:$KE = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$Equating potential energy and kinetic energy:$Mgh = \frac{3}{4} Mv^2$$v^2 = \frac{4gh}{3} \implies v = 2 \sqrt{\frac{gh}{3}}$Since $\omega = \frac{v}{R}$,the angular velocity is:$\omega = \frac{2}{R} \sqrt{\frac{gh}{3}}$

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. The angular velocity of the cylinder when it reaches the bottom of the plane will be

Similar Questions

$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is:

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?

$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

$A$ solid cylinder and a hollow cylinder,both of the same mass and same external diameter,are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

$A$ solid cylinder of mass $M$ and radius $R$ rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity $v$. What is the height of the inclined plane? (Assume $g$ is the acceleration due to gravity.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module