A body is rolling down an inclined plane. If kinetic energy of rotation is $40\%$ of translational kinetic energy, then the body is a
A body is rolling down an inclined plane. If kinetic energy of rotation is $40\%$ of translational kinetic energy, then the body is a
- A
Ring
- B
Cylinder
- C
Hollow ball
- D
Solid ball
Similar Questions
A disc of mass $m$ and radius $r$ is free to rotate about its centre as shown in the figure. A string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. The speed of the block as it descends through a height $h$, is .....
A disc of mass $m$ and radius $r$ is free to rotate about its centre as shown in the figure. A string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. The speed of the block as it descends through a height $h$, is .....
Two bodies have their moments of inertia $I$ and $2 I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio
Two bodies have their moments of inertia $I$ and $2 I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio
- [AIPMT 2005]
Two identical circular loops are moving with same kinetic energy one rolls $\&$ other slides. The ratio of their speed is
Two identical circular loops are moving with same kinetic energy one rolls $\&$ other slides. The ratio of their speed is
A hollow sphere of mass $m$ filled with a non-viscous liquid of same mass $m$ is released on a slope inclined at angle $q$ with the horizontal. The friction between the sphere and the slope is sufficient to prevent sliding and frictional forces between the inner surface of the sphere and the liquid is negligible. After descending a certain height ratio of translational and rotational kinetic energies is found to be $x:y$, find the numerical value of expression $(x+y)_{min}.$
A hollow sphere of mass $m$ filled with a non-viscous liquid of same mass $m$ is released on a slope inclined at angle $q$ with the horizontal. The friction between the sphere and the slope is sufficient to prevent sliding and frictional forces between the inner surface of the sphere and the liquid is negligible. After descending a certain height ratio of translational and rotational kinetic energies is found to be $x:y$, find the numerical value of expression $(x+y)_{min}.$
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]