Three solid spheres are made to move on a rou | vedclass

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(D) The frictional force always acts in the direction opposite to the tendency of motion of the contact point on the sphere relative to the surface.$1

Vedclass · Accepted Answer

Right,Left,Left

Vedclass · Answer

(D) The frictional force always acts in the direction opposite to the tendency of motion of the contact point on the sphere relative to the surface.$1$. For sphere $P$: It is given a clockwise spin. The contact point on the sphere has a tendency to move towards the left. Therefore,the friction force acts towards the right.$2$. For sphere $Q$: It is given a forward linear velocity $v$. The contact point on the sphere has a tendency to move towards the right. Therefore,the friction force acts towards the left.$3$. For sphere $R$: It is given a forward velocity $v$ and a counter-clockwise spin $\omega$. The velocity of the contact point is $v_{cp} = v - R\omega$. Since the sphere is rotating such that the bottom point moves left due to rotation and right due to translation,if $v > R\omega$,the net tendency of the contact point is to the right,so friction acts to the left. However,looking at the figure,the rotation is counter-clockwise,meaning the bottom point moves to the left due to rotation. If $v  R\omega$,friction is left. If $v < R\omega$,friction is right. Based on the standard interpretation of this specific problem,the correct sequence is Right,Left,Left.

Three solid spheres are made to move on a rough horizontal surface. Sphere $P$ is given a spin and released. Sphere $Q$ is given a forward linear velocity. Sphere $R$ is given linear and rotational motions as shown in the figure. Directions of the friction force on spheres $P, Q, R$ are respectively

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