A tennis ball is a hollow sphere with a thin wall. It is set rolling without sli

Question

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.15 m/s on a horizontal section of a track as shown in the figure below. It rolls around the inside of a vertical circular loop of radius r = 45.2 cm. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point h = 24.0 cm below the horizontal section. (a) Find the ball's speed at the top of the loop. (b) Is the ball going to fall from the top of the loop?

Explanation / Answer

(a)

From energy conservation,

K.E at bottom = P.E at top + K.E at top

(1/2)mv^2 + (1/2)lw2^2 + mgh2 = (1/2)mv1^2 + (1/2)lw1^2

(1/2)mv^2 + (1/2)*(2mr^2 / 3)*(v2^2 / r) + mgh2 = (1/2)mv1^2 + (1/2)*(2mr^2 / 3)*(v1^2 / r)

5v2^2 / 6 + gh2 = 5*v1^2 / 6

where v2 = 4.15 m/s

h2 = 2r = 0.904 m

so, v1 = 2.73 m/s

(b)

.lf the speed at the top is greater than sqrt(gr) then it will not fall from top and continue in circular path motion.

v' = sqrt (9.8*0.452)

v' = 2.10 m/s

v2 > v'

so, ball will not fall.

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