A solid marble of mass m = 20 kg and radius r = 3 cm will roll without slipping

Question

A solid marble of mass m = 20 kg and radius r = 3 cm will roll without slipping along the loop-the-loop track shown in the figure if it is released from rest somewhere on the straight section of track.The radius of the loop-the-loop is R = 1.20 m.

From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop?

If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?

Explanation / Answer

moment of inertia about the contact point. I=mr^2*2/5+mr^2=7mr^2/5. conservation of energy. mgh=mg*(2R-r)+I*w^2/2=mg*2R+7*m*v^2/10. so that g*(h-2R-r)=7v^2/10. vo v^2=10g(h-2R-r)/7. for it to be able to hold on to the track. v^2/(R-r)=g. so v^2=g*(R-r). so (R-r)=10*(h-2R-r)/7. so that h=3.25(m). ------------------------- assume that point Qs height is R(cant see the figure). so that mg*6R=mg*R+I*w^2/2. so that mg*5R=7mr^2w^2/10. centripetal force is the horizontal force. F=m*v^2/(R-r)=m*r^2*w^2/(R-r)=mg50R/7(R-r)=1436(N)

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