A hollow ball of mass m is released from rest at a height h above the base of a

Question

A hollow ball of mass m is released from rest at a height h above the base of a loop-the-loop track, as shown in the figure. The loop has a radius R = 22.0 cm. Assume that the ball?s radius is much smaller than the loop?s, and that the ball rolls without slipping at all times. Use g = 10 m/s^2.

What is the minimum value of h (in centimeters) necessary for the ball to make it all the way around the loop without losing contact with the track? (cm)

A hollow ball of mass m is released from rest at a height h above the base of a loop - the - loop track, as shown in the figure. The loop has a radius R = 22.0 cm. Assume that the ball?s radius is much smaller than the loop?s, and that the ball rolls without slipping at all times. Use g = 10 m/s^2. What is the minimum value of h (in centimeters) necessary for the ball to make it all the way around the loop without losing contact with the track? (cm)

Explanation / Answer

At the top of the loop, the ball will be experiencing a downward force from gravity, if this is not sufficient to provide a large enough force to cause the centripetal acceleration of the ball, then it will also experience a downward normal force from the track which will make up for the deficiency. Knowing this, we can see that the minimum velocity is when the normal force is 0, so the centripetal force is equal to the downward force of gravity. Centripetal Force is F = mV^2/R Force from gravity is mg. Since these are equal we have the equation: mV^2/R = mg V^2/R = g V^2 = Rg V=sqrt(Rg) At the top of the loop, the y-displacement of the ball will be h-2R. Since vf^2 = vi^2 + 2*a*d, we can plug in what we know and solve for d. Rg = 2gd R=2d d = R/2 Since d=h-2R and d=R/2, h-2R = R/2. h= 5R / 2

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