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 4.13 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 = 48.0 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 = 16.0 cm below the horizontal section. (a) Find the ball's speed at the top of the loop. (c) Find its speed as it leaves the track at the bottom.

Explanation / Answer

A. mghi + 0.5*mvi^2 + 0.5*I*wi^2 = mghf + 0.5*mvf^2 + 0.5*I*wf^2

Vf = [{2*9.8*m*(0.16 - 1.12) + (1 + 0.66*m*0.48^2/(0.5*m*0.48^2))4.13^2}/{(1 + 0.66*m*0.48^2/(0.5*m*0.48^2)}]^0.5 = 2.98 m/s

B. energy conservation

(KE + UE + RE)i = (KE + UE +RE)t
0.5mvi^2 + mghi + 0.5*I*wi^2 = 0.5mvt^2 + mght + 0.5*I*wt^2

0.5mvi^2 + mghi + 0.5*(0.66m*R^2)(vi/R)^2 = 0.5mvt^2 + mght + 0.5*(0.66m*R^2)(vt/R)^2

hi = r ; ht = -h + r

(5/6)*vt^2 = (5/6)*vi^2 + gh
vt = [4.13^2 + 6*9.8*0.16/5]^0.5 = 4.35

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