A solid marble of mass m = 10 kg and radius r = 2 cm will roll without slipping

Question

A solid marble of mass m = 10 kg and radius r = 2 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 = 1.35 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? (Use conservation of mechanical energy (potential, linear and rotational kinetic energy) and set the centripetal force equal to the weight to find the minimun height.)

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?

A solid marble of mass m = 10 kg and radius r = 2 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 = 1.35 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? (Use conservation of mechanical energy (potential, linear and rotational kinetic energy) and set the centripetal force equal to the weight to find the minimun height.) 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

Total energy at the start is TE = mgH; where H = ? m, the min. height you're looking for, m = 10 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 angular kinetic energy. pe is the reduced potential energy at h < H, which is the diameter of the loop so h = 2R = 2.7 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*1.35 = ? m

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