Problem 17 - Entrance Test

A uniform solid sphere of mass M and radius R is rotating with an angular speed ω about an axis passing through its center. If the sphere is placed gently on a rough horizontal surface, it will eventually roll without slipping. What is the final linear speed of the center of mass of the sphere when it begins to roll without slipping?

A. A) (2/7)RωB. B) (5/7)RωC. C) (2/5)RωD. D) (7/5)RωE. E) (1/2)Rω

Correct: A

When the sphere is placed on the rough surface, friction will act to change both its translational and rotational motion until the condition for rolling without slipping (v_CM = Rω_f) is met. There is no external horizontal force on the system (sphere + surface) other than friction, which is an internal force when considering the sphere and the Earth. If we consider the system as only the sphere, friction is an external force, but it does not exert a torque about the point of contact in a specific frame. Let's use angular momentum conservation about the point of contact. Consider the angular momentum about the instantaneous point of contact with the ground. At the moment the sphere is placed down, the point of contact instantaneously becomes the pivot point for calculating angular momentum change. The friction force at the bottom of the sphere (which starts out slipping) will create a torque about the center of mass. However, there is no *net* external torque about the point of contact with the ground in the horizontal direction (the friction force acts *through* the point of contact, and the normal force/gravity pass through a fixed line if the surface is flat). Therefore, angular momentum is conserved about the point of contact with the ground, or equivalently, if the friction does not produce external torque about the initial pivot point for calculation (which might be complicated). A simpler way is to consider the linear impulse and angular impulse. Let's consider the change in angular momentum about the point of contact P. The friction force f_k acts at P. No torque is exerted by f_k about P. Gravity and Normal force pass through the CM, so they do not exert torques about P relative to the CM's vertical line. Initial state: Linear speed v₀ = 0 Angular speed = ω Angular momentum about P (L_P, initial) = I_CM * ω + Mv_CM * R_perp Since v_CM = 0, L_P, initial = I_CM * ω = (2/5)MR²ω (for a solid sphere). Final state (rolling without slipping): Linear speed = v_f Angular speed = ω_f = v_f / R Angular momentum about P (L_P, final) = I_CM * ω_f + Mv_f * R I_CM * ω_f = (2/5)MR² * (v_f/R) = (2/5)MRv_f Mv_f * R = Mv_f * R So, L_P, final = (2/5)MRv_f + MRv_f = (7/5)MRv_f By conservation of angular momentum about the contact point P: L_P, initial = L_P, final (2/5)MR²ω = (7/5)MRv_f Divide both sides by (MR): (2/5)Rω = (7/5)v_f v_f = (2/5)Rω * (5/7) v_f = (2/7)Rω The correct choice is A.