A tennis ball is a hollow sphere with a thin wall. It is set rolling without sli
ID: 1418384 • Letter: A
Question
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.11 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 = 45.6 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 = 22.0 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.
This answer has not been graded yet.
(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?
higherlower the sameThe ball never makes it to the top of the loop.
(e) Explain your answer.
Explanation / Answer
a)
Apply conservation of energy
0.5*m*v1^2 + 0.5*I*w1^2 = 0.5*m*v2^2 + 0.5*I*w2^2 + m*g*H
0.5*m*v1^2 + 0.5*(2/3)*m*R^2*w1^2 = 0.5*m*v2^2 + 0.5*(2/3)m*R^2*w2^2 + m*g*2*r
0.5*m*v1^2 + (1/3)*m*(R*w1)^2 = 0.5*m*v2^2 + (1/3)*m*(R*w2)^2 + m*g*2*r
0.833*m*v1^2 = 0.833*m*v2^2 + 2*m*g*r
v2 = sqrt(v1^2 - 2*g*r/0.833)
= sqrt(4.11^2 - 2*9.8*0.456/0.833)
= 2.48 m/s
b) minimum speed of ball at the top of the ball, v_min = sqrt(g*r)
= sqrt(9.8*0.456)
= 2.11 m/s
2.48 m/s > 2.11 m/s
so, the ball will not fall down
c) v_bottom = sqrt(v1^2 + g*h/0.833)
= sqrt(4.11^2 + 2*9.8*0.22)
= 4.6 m/s
d) The ball never makes it to the top of the loop.
e) the ball do not have enough kinetic energy to move to the top of the loop.
v2^2 - v1^2 = 2*g*H
v2 = sqrt(v1^2 + 2*g*H)
= sqrt(4.11^2 + 2*(-9.8)*2*0.456)
= indeterminate
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.