NOTE: the center of mass of the ball initially is (h + r) above the bottom of th

Question

NOTE: the center of mass of the ball initially is (h + r) above the bottom of the loop. As it passes the top of the loop, the center of mass is (2R ? r) above the bottom on the loop. A small, solid sphere of mass 0.9 kg and radius 54 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 4.85 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 54 cm. What is the minimum value of h such that the sphere completes the loop? The acceleration due to gravity is 9.8 m/s2 . The moment of inertia for a solid sphere is 2/5mr^2 .Answer in units of m.

Explanation / Answer

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The center of mass of the ball initially

is (h + r) above the bottom of the loop. As it

passes the top of the loop, the center of mass

of the ball is (2R - r) above the bottom of the

loop.

A small, solid sphere of mass 0.9 kg and

radius 29 cm rolls without slipping along the

track consisting of slope and loop-the-loop

with radius 2 m at the end of the slope. It

starts from rest near the top of the track at a

height h, where h is large compared to 29 cm.

What is the minimum value of h (in terms

of the radius of the loop R) such that the

sphere completes the loop? The acceleration

due to gravity is 9.8 m/s^2

The moment of inertia for a solid sphere is 2/5 m r^2

Answer

Ei = Ef

m g (h + r) = m g( 2 R - r) + 1/2 mv^2 + 1/2 I w^2

on top going in a circle of radius R-r

m g = mv^2/( R - r)

mv^2 = m g ( R - r)

1/2I w^2 = 1/5 m r^2 w^2 = 1/5 m v^2

so
mg ( h + r) = m g ( 2 R - r) + (1/2 + 1/5) m g ( R - r)

mg cancels

h + .29 = (2*2 - .29) + (1/2 + 1/5) (2-.29)

h=4.62 m

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