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 4.14 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 = 49.0 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 = 18.0 cm below the horizontal section.

(a) Find the ball's speed at the top of the loop.
m/s

(b) Demonstrate that the ball will not fall from the track at the top of the loop.

(c) Find its speed as it leaves the track at the bottom.
m/s

(d) Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop?

higher, lower , the sameThe ball never makes it to the top of the loop.

(e) Explain your answer.

Explanation / Answer

a) Since it is rolling the kinetic energy will be written like this,

KE = 0.5*m*v^2 + 0.5*I*w^2, where I = (2/3)*m*r^2 and w = v*r so KE = (5/6)*m*v^2.

The intital kinetic energy = (5/6)*m*v^2 = (5/6)*m*4.14^2 = 14.283*m.

The final kinetic energy = (5/6)*m*v^2.

The work done by gravity = mgh = m*9.8*2*r = m*9.8*0.49 = 4.8*m.

So the change in kE = 14.283*m-0.833*m*v^2

but the work done equals to change in kinetic energy

S0 4.8*m = 14.283*m-0.833*m*v^2 => v = 3.37 m/s.

b) The acceleration towards the center at the top, = v^2/r = 23.23 m/s^2. So the force vertically downwards will be F = m*23.23.

or N+mg = m*23.23 or N = 22.23*m which is positive. Since the normal exits that means there will be contact between the two surfaces or it will not fall.

c) The work done in this case, W = mgh = 9.8*0.18*m = 1.764*m J.

The initial KE = 14.283*m, the final KE = (5/6)*m*v^2. so by work energy principle,

1.764*m = (5/6)*m*v^2-14.283 => v = 4.388 m/s

d) the KE = 0.5*m*v^2

initial KE = 0.5*m*4.14^2 = 8.5698*m. final kE = 0.5*m*v^2. The work done will be = 4.8*m

So from work energy principle, 4.8 m = 8.5698*m-0.5*m*v^2 = > v = 2.74 m/s.which is lower.

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