A sphere of mass M and radius r slips on a ro | vedclass

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

Question

(D) The angular momentum of the sphere about the point of contact with the ground remains conserved because the frictional force acts through the poin

Vedclass · Accepted Answer

$\frac{6V_0}{7}$

Vedclass · Answer

(D) The angular momentum of the sphere about the point of contact with the ground remains conserved because the frictional force acts through the point of contact,exerting no torque about it.Initial angular momentum about the point of contact:$L_i = M V_0 r + I_{cm} \omega = M V_0 r + (\frac{2}{5} M r^2) (\frac{V_0}{2r}) = M V_0 r + \frac{1}{5} M V_0 r = \frac{6}{5} M V_0 r$Final angular momentum when pure rolling starts (let final velocity be $v'$ and angular velocity be $\omega' = \frac{v'}{r}$):$L_f = M v' r + I_{cm} \omega' = M v' r + (\frac{2}{5} M r^2) (\frac{v'}{r}) = M v' r + \frac{2}{5} M v' r = \frac{7}{5} M v' r$Equating $L_i = L_f$:$\frac{6}{5} M V_0 r = \frac{7}{5} M v' r$$v' = \frac{6}{7} V_0$

$A$ sphere of mass $M$ and radius $r$ slips on a rough horizontal plane. At some instant,it has translational velocity $V_0$ and rotational velocity about the centre $\frac{V_0}{2r}$. The translational velocity when the sphere starts pure rolling motion is

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 $1 \ kg$ is rolling on a rough table with a linear speed of $1 \ m/s$. Its total kinetic energy is ....... $J$.

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300 \, J$. The translational kinetic energy of the disc is:

Which of the following statements is incorrect for a spherical body rolling without slipping on a rough horizontal ground at rest?

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $4 \ J$,then the total kinetic energy of the disc is: (in $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module