@0 100% 6:03 PM elearning.wmich.edu Sprint LTE Question 3 (1 point) 1. A tennis

Question

@0 100% 6:03 PM elearning.wmich.edu Sprint LTE Question 3 (1 point) 1. A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 411 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 = 50.0 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 = 17.0 cm below the horizontal section. Find the ball's speed at the top of the loop a) 3.31 m/s

Explanation / Answer

let u = 4.11 m/s, v = ?

Apply conservation of mechanical energy

(1/2)*m*u^2 + (1/2)*I*w^2 = (1/2)*m*v^2 + (1/2)*I*w'^2 + m*g*H

(1/2)*m*u^2 + (1/2)*(2/3)*m*R^2*w^2 = (1/2)*m*v^2 + (1/2)*(2/3)*m*R^2*w'^2 + m*g*H

(1/2)*m*u^2 + (1/3)*m*R^2*w^2 = (1/2)*m*v^2 + (1/3)*m*R^2*w'^2 + m*g*H

(1/2)*m*u^2 + (1/3)*m*u^2 = (1/2)*m*v^2 + (1/3)*m*v^2 + m*g*H (since u = R*w and v = R*w')

(5/6)*m*u^2 = (5/6)*m*v^2 + m*g*2*r

(5/6)*u^2 = (5/6)*v^2 + 2*g*r

u^2 = v^2 + (12/5)*g*r

v = sqrt(u^2 - 2.4*g*r)

= sqrt(4.11^2 - 2.4*9.8*0.5)

= 2.46 m/s

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