A solid sphere with a diameter of 0.17 m is released from rest; it then rolls wi

Question

A solid sphere with a diameter of 0.17 m is released from rest; it then rolls without slipping down a ramp, dropping through a vertical height of 0.61 m. The ball leaves the bottom of the ramp, which is 1.22 m above the floor, moving horizontally. a) Through what horizontal distance d does the ball move before landing? b) How many revolutions does the ball make during its fall? c) If the ramp were to be made frictionless, would the distance d increase, decrease, or stay the same? Explain. (a: 1.5 m, b: 2.7 rev, c: decrease (why?) ) A solid sphere with a diameter of 0.17 m is released from rest; it then rolls without slipping down a ramp, dropping through a vertical height of 0.61 m. The ball leaves the bottom of the ramp, which is 1.22 m above the floor, moving horizontally. a) Through what horizontal distance d does the ball move before landing? b) How many revolutions does the ball make during its fall? c) If the ramp were to be made frictionless, would the distance d increase, decrease, or stay the same? Explain. (a: 1.5 m, b: 2.7 rev, c: decrease (why?) )

Explanation / Answer

(A) Applying energy conservation to find its speed at moment it leaves the bottom of ramp.

m gh = (m v^2/ 2) + ( I w^2 / 2)

m g h = m v^2 /2 + ( 2 m r^2 /5) (v / r)^2 / 2

m g h = m v^2 /2 + m v^2 / 5 = 7 m v^2 / 10

9.8 x 0.61 = 7 v^2 / 10

v = 2.92 m/s

it leaves horizontally, hence initial vertical velocity will be zero.

y = v0y t + a t^2 / 2

- 1.22 = 0 - 9.8 t^2 / 2

t = 0.50 sec

in horizontal (a = 0 )

so D = v0 t = 2.92 x 0.50 = 1.46 m

(b) w = v /r = 2.92 / (0.17/2) = 34.35 rad/s

theta = w t = 34.35 x 0.5 = 17.17 rad

revolutions = theta / 2pi = 2.72 revolutions

(c) if ramp is frictionless then it will get only tranlastional KE. (No rotational KE)

so its velocity wll be greater in this case.

t depends on height, so it wil remain same.

v0 -> increases, and D = v0 t

so D -> increases

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