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

Two identical circular loops are moving with the same kinetic energy; one rolls and the other slides. The ratio of their speed is:

$A$ sphere rolls without slipping. The radius of gyration about an axis passing through its center of mass is $K$. If the radius of the sphere is $R$,what fraction of the total energy is in the form of rotational kinetic energy?

$A$ solid homogeneous sphere of mass $M$ and radius $r$ is moving on a rough horizontal surface,partly rolling and partly sliding. During this kind of motion of this sphere,

$A$ ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If the radius of the ball is $R$,then the fraction of total energy associated with its rotational energy is:

$A$ wheel of radius $1 \text{ m}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

Vedclass Test Series

Exam Paper Generator

Online Exam Module