A thin-walled, hollow spherical shell of mass m and radius r starts from rest an

Question

A thin-walled, hollow spherical shell of mass m and radius r starts from rest and rolls without slipping down the track shown in the figure. Points A and B are on a circular part of the track having radius R. The diameter of the shell is very small compared to h0 and R are, and rolling
friction is negligible.

a) what is the minimum height h0 for which this shell will make a complete loop-the-loop on the circular part of the track?

b) how hard does the track push on the shell at point B, which is at the same level as the center of the circle?

c) Suppose the track had no friction and the shell was released from the same height h0 you found in part (a). Would it make a complete loop-the-loop. How do you know?

d) In part (c) how hard does the track push on the shell at point A, the top of the circle? How hard did it push on the shell in part (a)

Explanation / Answer

A) To make a complete loop, you must have a minimum velocity such that the centripetal acceleration is at least as great as gravity: at the threshold, v²/R = g, or v² = gR
For a "thin-walled, hollow spherical shell",

I = (2/3)mr², so the translational and rotational KEs sum to
KE = ½mv² + ½I² = ½mv² + ½(2/3)mr²(v/r)² = (5/6)mv²,
so at the top of the loop, the (minimum) total energy
E = PE + KE = mg(2R) + (5/6)m(gR) = (17/6)mgR

Conclusion: At the release point, where all of the energy is potential, the height must be h = 17R/6.

B) At B, the height is R, so the KE must be (17/6 - 1)mgR. Then
(11/6)mgR = (5/6)mv²
(11/5)mg = mv²/R
The right hand side is the centripetal force Fc; hence
Fc = (11/5)mg

C) If the shell (or any other object) is sliding, our required energy is
E = PE + KE = mg(2R) + ½m(gR) = (15/6)mgR
in order to stay on the loop. We have more energy than that, so the sliding object DOES complete the loop.

D) At A, the height is 2R, so the KE must be (17/6 - 2)mgR. Then
(5/6)mgR = ½mv² multiply through by 2/R
(5/3)mg = mv²/R
To find the net force on the track, we need to subtract the weight.
net F = (5/3)mg - mg = (2/3)mg

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