A solid marble of mass m = 45 kg and radius r = 7 cm will roll without slipping

Question

A solid marble of mass m = 45 kg and radius r = 7 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 = 1.00 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?
2.70 m

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

I have already answered this question previously.

Total energy at the start is TE =mgH; where H = ? m, the min. height you're looking for, m = 35 kg (a really really massive marble), and g is g.

At the top of the loop, TE = ke + pe = 1/2 mv^2(1 + k) + mgh; where k = 2/5 for a solid sphere and assuming no slippage. ke includes both linear and angularkinetic energy. pe is the reducedpotential energyat h < H, which is the diameter of the loop so h = 2R = 2.6 m.

From the conservation of energy, TE = mgH = 1/2 mv^2(1 + k) + mgh = TE; solve for H.

H = (1/2 v^2(1 + k) + gh)/g Now we need v so as g = a = v^2/R is true, the force of gravity is offset by the centrifugal force. Then,

H = (1/2 gR(1 + k) + gh)/g = 1/2 R(1 + 2/5) + 2R = 1/2*7R/5 + 2R = R(7/10 + 20/10) = (27/10)R = 2.7R. Thus the min height H for any loop radius R must be 2.7 times the loop radius to stay on the track when at the top of the loop h. H = 2.7*1.3 = 3.51 m ANS.

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