If $L, M$ and $P$ are the angular momentum, mass and linear momentum of a particle respectively which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$
If $L, M$ and $P$ are the angular momentum, mass and linear momentum of a particle respectively which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$
- A
$\frac{{{L^2}}}{{2M}}$
- B
$\frac{{{P^2}}}{{2MR}}$
- C
$\frac{{{L^2}}}{{2M{R^2}}}$
- D
$\frac{{MP}}{2}$
Similar Questions
A solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........cm s ^{-1}$.
A solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........cm s ^{-1}$.
- [JEE MAIN 2023]
A sphere of mass $50\,gm$ and diameter $20\,cm$ rolls without slipping with a velocity of $5\,cm/sec$ . Its total kinetic energy is
A sphere of mass $50\,gm$ and diameter $20\,cm$ rolls without slipping with a velocity of $5\,cm/sec$ . Its total kinetic energy is
When a body is rolling without slipping on a rough horizontal surface, the work done by friction is ........
When a body is rolling without slipping on a rough horizontal surface, the work done by friction is ........
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 ($2$)
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 ($2$)
- [IIT 2017]
A $2 \,{kg}$ steel rod of length $0.6\, {m}$ is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \,{ms}^{-1}$. (Take $g = 10\, {ms}^{-2}$ )
A $2 \,{kg}$ steel rod of length $0.6\, {m}$ is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \,{ms}^{-1}$. (Take $g = 10\, {ms}^{-2}$ )
- [JEE MAIN 2021]