A thin hollow cylinder,open at both ends,slid | vedclass

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

Question

(B) Case $1$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{slide} = \frac{1}{2}mv^2$Case $2$: W

Vedclass · Accepted Answer

$1 : 2$

Vedclass · Answer

(B) Case $1$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{slide} = \frac{1}{2}mv^2$Case $2$: When it rolls without slipping,it possesses both translational and rotational kinetic energy.The total kinetic energy is given by $K_{roll} = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.For a thin hollow cylinder,the moment of inertia $I = mR^2$,so $mk^2 = mR^2$,which implies $\frac{k^2}{R^2} = 1$.Substituting this into the formula: $K_{roll} = \frac{1}{2}mv^2(1 + 1) = mv^2$.Now,the ratio of kinetic energies is:$\frac{K_{slide}}{K_{roll}} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$.Thus,the ratio is $1 : 2$.

$A$ thin hollow cylinder,open at both ends,slides without rotating and then rolls without slipping with the same speed. The ratio of kinetic energies in the two cases will be:

Similar Questions

$A$ solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is ................

$A$ solid circular disc of mass $50 \,kg$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \,m/s$. The absolute value of work done on the disc to stop it is . . . . . . $J$.

$A$ sphere of mass $0.5 \ kg$ and diameter $1 \ m$ is rolling without slipping with a constant velocity of $5 \ m/s$. The ratio of its rotational kinetic energy to its total kinetic energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module