Solid spherical ball is rolling on a friction | vedclass

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

Question

$K _{	ext {total }}= K _{	ext {rotational }}+ K _{	ext {Translational }}$

$K _{ total }=\frac{1}{2} I _{ cm } \omega^{2}+\frac{1}{2} mV _{ cm }^

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

$K _{	ext {total }}= K _{	ext {rotational }}+ K _{	ext {Translational }}$

$K _{ total }=\frac{1}{2} I _{ cm } \omega^{2}+\frac{1}{2} mV _{ cm }^{2}$

$v _{ cm }= R \omega$ for pure rolling

$I _{ cm }=\frac{2}{5} mR ^{2}$

$K _{ Rot }=\frac{1}{2} I _{ cm } \omega^{2}=\frac{1}{2} 	imes \frac{2}{5} mR ^{2} 	imes \frac{ v _{ cm }^{2}}{ R ^{2}}=\frac{1}{5} mv _{ cm }^{2}$

$K _{	ext {Total }}=\frac{1}{5} mv _{ cm }^{2}+\frac{1}{2} mv _{ cm }^{2}=\frac{7}{10} mv _{ cm }^{2}$

$\frac{ K _{ Rot }}{ K _{	ext {Total }}} \frac{\frac{1}{5} mv _{ cm }^{2}}{\frac{7}{10} mv _{ cm }^{2}}=\frac{2}{7}$

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 ................

Similar Questions

Two disc one of density $7.2\, g/cm^3$ and the other of density $8.9\, g/cm^3$ are of same mass and thickness. Their moments of inertia are in the ratio

In a gravity free space, a man of mass $M$ standing at a height $h$ above the floor, throws a ball of mass $m$ straight down with a speed $u$ . When the ball reaches the floor, the distance of the man above the floor will be

A uniform rod of mass $m$ and length $l$ rotates in a horizontal plane with an angular velocity $\omega $ about a vertical axis passing through one end. The tension in the rod at a distance $x$ from the axis is

We have two spheres, one of which is hollow shell and the other solid. They have identical masses and moment of inertia about their respective diameters. The ratio of their radius is given by