A uniform ring of radius R is moving on a horizontal surface with speed v , then climbs up a ramp of

English

Gujarati

Hindi

6.System of Particles and Rotational Motion

medium

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

A

$v^{2} / 2 g$

B

$v^{2} / g$

C

$3 v^{2} / 2 g$

D

$2 v^{2} / g$

(KVPY-2016)

Solution

$(b)$ 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 $v=R \omega$

$\therefore \frac{1}{2} m v^{2}+\frac{1}{2} m R^{2} \times \frac{v^{2}}{R^{2}} =m g h$

$\Rightarrow m v^{2} =m g h$

$\Rightarrow h =\frac{v^{2}}{g}$

Standard 11

Physics

Share

Similar Questions

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \;kg – m ^{2}$ and $10\; rad \,s^{-1}$ respectively while those for the second one are $0.2 \;kg – m ^{2}$ and $5\; rad \,s ^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is ………..$J$

hard

(JEE MAIN-2020)

One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure $1$ . In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $\mathrm{r}$. The finger rotates with an angular velocity $\omega_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure $2$). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.

(IMAGE)

($1$) The total kinetic energy of the ring is

$[A]$ $\mathrm{M} \omega_0^2 \mathrm{R}^2$ $[B]$ $\frac{1}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$ $[C]$ $\mathrm{M \omega}_0^2(\mathrm{R}-\mathrm{r})^2$ $[D]$ $\frac{3}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$

($2$) The minimum value of $\omega_0$ below which the ring will drop down is

$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$ $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$ $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$ $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$

Givin the answer quetion ($1$) and ($2$)

normal

(IIT-2017)

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent)

normal

An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $………….\,kg – m ^2$

easy

(NEET-2022)

Write the formula for power and angular momentum in rotational motion.

medium