A solid sphere is released from the top of a ramp that is at a height h 1 = 2.50

Question

A solid sphere is released from the top of a ramp that is at a height h1 = 2.50 m. It goes down the ramp, the bottom of which is at a height of h2 = 2.05 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.18 m.

A solid sphere is released from the top of a ramp that is at a height h1 = 2.50 m. It goes down the ramp, the bottom of which is at a height of h2 = 2.05 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.18 m. Through what horizontal distance d does the ball travel before landing? How many revolutions does the ball make during its fall?

Explanation / Answer

R = 0.18/2 = 0.09 m
Moment of intertia I = 2/5 * m R^2

Let v1 = speed of the sphere at the bottom of the ramp.
By conservation of energy,
m g h1 = 1/2 * m * v1^2 + 1/2 * I * w^2
m g h1 = 1/2 * m * v1^2 + 1/2 * 2/5 * m R^2 * v1^2/R^2
m g h1 = 1/2 * m * v1^2 + 1/5 * m * v1^2
m g h1 = 0.7 * m * v1^2
v1 = sqrt(g h1/0.7) = sqrt(9.8 * 2.05/0.7)
v1 = 5.357m/s

Let w1 = angular velocity at the bottom of the ramp
w1 = v1/R =5.357/0.09
w1 = 59.522 rad/s
Time to fall t = sqrt(2 * h2/g) = sqrt(2 * 2.50/9.8) = 0.714 s

Angular displacement during fall = w1 * t
= 59.522 * 0.714 rad
= (59.522 * 0.714)/(2 pi) revolutions
= 6.76 revolutions

d = V1*t

d=5.357*0.714

=3.82 m

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