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

Question

h1 = 1.95 m.
It goes down the ramp, the bottom of which is at a height of
h2 = 1.38 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 0.14 m.

(a) Through what horizontal distance d does the ball travel before landing?

(b) How many revolutions does the ball make during its fall?

Answer 1

Rball = .07 meter

KE = (1/2) m v^2 + (1/2)I w^2
I = (2/5) m r^2
v = w Rball = .07 w
so
Ke = (1/2) m (.07 w)^2 + (2/5) m .07^2 w^2

= m (9/10) (.0049) w^2 = .00441 m w^2

Change in Pe down ramp = m g h = m(9.81)(.57) = .55917 m

Ke = change in Pe
.55917 m = .00441 m w^2
w^2 = 127
w = 11.3 radians/second
v = w r = .07*11.3 = .788 meters/second horizontal speed
Now how long to fall the last 1.38 meters?
1.38 = (1/2) g t^2
t^2 = .281
t = .53 seconds in the air
how far in .53 seconds at .788 m/s
x = v t = .788 * .53 = .418 meters

radians/second = 11.3
for .53 seconds = 5.99 radians
5.99/2 pi = .953 revolutions

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