A solid marble of mass m = 50 kg and radius r = 8 cm will roll without slipping

Question

A solid marble of mass m = 50 kg and radius r = 8 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 = 1.05 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?
2.83 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

In order for the ball to remain on the track at the top of the loop it must be going fast enough for the centrifugal force at that point to balance gravity. Let that velocity be V. The centrifugal force is then m*V²/R, which must equal m*g. Solve for V² to get

V² = R*g

The kinetic energy of the ball at that point is 0.7*m*V² (the 0.7 factor includes linear and rotational kinetic energy). This energy comes from the initial potential energy at the top of the track, and this is m*g*y, where y is the change in height between the two points. h = h - 2*R

Then 0.7*m*V² = m*g*( h - 2*R)

and substituting for V²:

0.7*R*g= g*( h - 2*R)

0.7*R = h - 2*R

h = 2.7*R

h = 2.7*1.05 =2.835m

(b) Find the velocity at point Q from equating m*g*y to 0.7*m*V²

V² = g*y/0.7

horizontal force component is m*V²/R = m*g*y/(0.7*R). In this case y = 6R - R = 5*R, so

Fh = m*g*(5/0.7)

Fh = 50*9.8*(7.142) = 3500 N

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