Write the formula of work done by torque in rotational rigid body about a the fixed axis.
Write the formula of work done by torque in rotational rigid body about a the fixed axis.
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]
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]
An air compressor is powered by a $200\,rad\,s^{-1}$ electric motor using a $V-$ belt drive. The motor pulley is $8\,cm$ in radius, and the tension in the $V-$ belt is $135\,N$ on one side and $45\,N$ on the other. The power of the motor will be ...... $kW$.
An air compressor is powered by a $200\,rad\,s^{-1}$ electric motor using a $V-$ belt drive. The motor pulley is $8\,cm$ in radius, and the tension in the $V-$ belt is $135\,N$ on one side and $45\,N$ on the other. The power of the motor will be ...... $kW$.
Four point masses are fastened to the corners of $a$ frame of negligible mass lying in the $xy$ plane. Let $w$ be the angular speed of rotation. Then
Four point masses are fastened to the corners of $a$ frame of negligible mass lying in the $xy$ plane. Let $w$ be the angular speed of rotation. Then
$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of $M/m =?$
$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of $M/m =?$