A solid sphere is released from the top of a ramp that is at a height A solid sp

Question

A solid sphere is released from the top of a ramp that is at a height

A solid sphere is released from the top of a ramp that is at a height h1 = 1.70 m. It goes down the ramp, the bottom of which is at a height of h2 = 1.25 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

Rball = .09 meter

KE = (1/2) m v^2 + (1/2)I w^2
I = (2/5) m r^2
v = w Rball = .09 w
so
Ke = (1/2) m (.09 w)^2 + (2/5) m .09^2 w^2

= m (9/10) (.0081) w^2 = .00729 m w^2

Change in Pe down ramp = m g h = m(9.81)(.45) = .44145 m

Ke = change in Pe
.44145 m = .00729 m w^2

w = 24.61 radians/second
v = w r = .09*24.61 = 2.215 meters/second horizontal speed
Now how long to fall the last 1.25 meters?
1.25 = (1/2) g t^2

t = .505 seconds in the air
how far in .505 seconds at 2.215 m/s
x = v t = 2.215 * .505 = 1.12 meters

radians/second = 24.61
for .505 seconds = 12.43 radians
12.43/2 pi = 1.98 revolutions

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