A uniform ring of radius R is moving on a hor | vedclass

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

Question

$(b)$  As ring is rolling without any slippage, Total initial kinetic energy $=$ Total final potential energy at height $h$

$\Rightarrow$ Kinc

Vedclass · Accepted Answer

$v^{2} / g$

Vedclass · Answer

$(b)$  As ring is rolling without any slippage, Total initial kinetic energy $=$ Total final potential energy at height $h$

$\Rightarrow$ Kinctic energy of translation $+$ Kinctic energy of rotation = Potential energy of ring at height $h$

$\Rightarrow \quad \frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h$

For ring, $I=m R^{2}$ and $v=R \omega$

$	herefore \frac{1}{2} m v^{2}+\frac{1}{2} m R^{2} 	imes \frac{v^{2}}{R^{2}} =m g h$

$\Rightarrow m v^{2} =m g h$

$\Rightarrow h =\frac{v^{2}}{g}$

A uniform ring of radius $R$ is moving on a horizontal surface with speed $v$, then climbs up a ramp of inclination $30^{\circ}$ to a height $h$. There is no slipping in the entire motion. Then, $h$ is

Similar Questions

A wheel is rotating with an angular speed of $20\,rad/sec$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg-m^2$, then the work done by the torque in two seconds will be .......... $J$

A thin uniform rod oflength $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end . Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

Write the formula for rotational kinetic energy.

A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the ratio of their velocities will be