A ring of mass m is rolling without slipping | vedclass

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

Question

(C) The total kinetic energy of the system is the sum of the kinetic energy of the ring and the kinetic energy of the rod.For rolling without slipping

Vedclass · Accepted Answer

$\frac{5}{3}mv^2$

Vedclass · Answer

(C) The total kinetic energy of the system is the sum of the kinetic energy of the ring and the kinetic energy of the rod.For rolling without slipping,$\omega = \frac{v}{R}$.Kinetic energy of the ring $(KE_{ring})$: $KE_{ring} = \frac{1}{2}mv^2 + \frac{1}{2}I_{ring}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Kinetic energy of the rod $(KE_{rod})$: The rod rotates about the center of the ring. The moment of inertia of a rod of length $2R$ about its center is $I_{rod} = \frac{m(2R)^2}{12} = \frac{4mR^2}{12} = \frac{1}{3}mR^2$.$KE_{rod} = \frac{1}{2}mv^2 + \frac{1}{2}I_{rod}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{3}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{6}mv^2 = \frac{3+1}{6}mv^2 = \frac{4}{6}mv^2 = \frac{2}{3}mv^2$.Total kinetic energy = $KE_{ring} + KE_{rod} = mv^2 + \frac{2}{3}mv^2 = \frac{5}{3}mv^2$.

$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown in the figure. $A$ rod of identical mass $m$ is fixed along one of its diameters. The total kinetic energy of the system is:

Similar Questions

$A$ hollow sphere of mass $2 \, kg$ and diameter $0.2 \, m$ is rolling on an inclined plane with a velocity of $0.5 \, m/s$. The kinetic energy of the sphere will be .......... $J$.

$A$ wheel is rolling on the ground with a speed of $2 \ m/s$. What is the velocity of the endpoints of the horizontal diameter of the wheel?

Find the displacement of the point of contact with the ground when a ring of radius $R$ completes a half-rotation.

Three solid spheres are made to move on a rough horizontal surface. Sphere $P$ is given a spin and released. Sphere $Q$ is given a forward linear velocity. Sphere $R$ is given linear and rotational motions as shown in the figure. Directions of the friction force on spheres $P, Q, R$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module