A solid sphere is released from the top of a ramp that is at a height h1 = 2.35

Question

A solid sphere is released from the top of a ramp that is at a height h1 = 2.35 m. It rolls down the ramp without slipping. The bottom of the ramp is at a height of h2 = 1.8 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 0.12 m. A ball begins at a height labeled h_1, rolls down a ramp to a lower height labeled h_2, at which point it falls in a curved arc down to the ground where it lands a distance labeled d from the end of the ramp. (a) Through what horizontal distance d does the ball travel before landing? Consider the motion along the ramp first. What is the speed of the ball at the bottom of the ramp? Now consider the motion of the ball after it leaves the ramp. What is the path of the ball once it leaves the ramp? m (b) How many revolutions does the ball make during its fall? rev

Explanation / Answer

a)

using conservation of energy

Potential energy at the top of ramp = kinetic energy at the bottom + rotational KE + PE at the bottom of ramp

mg h1 = (0.5) m v2 + (0.5) I w2 + mg h2

mg h1 = (0.5) m v2 + (0.5) (0.4) (m r2) (v/r)2 + mg h2

mg (h1 - h2 ) = (0.7) m v2

(9.8) (2.35 - 1.8) = (0.7) v2

v = 2.8 m/s

consider the motion of the ball after falling

Y = vertical displacement = h2 = 1.8 m

a = acceleration = 9.8

Voy = initial velocity in Y-direction = 0 m/s

t = time taken to hit the ground

using the equation

Y = Voy t + (0.5) a t2

1.8 = 0 t + (0.5) (9.8) t2

t = 0.61 sec

consider the motion along the X-direction

Vox = velocity along X-direction = v = 2.8 m/s

t = 0.61 sec

distance travelled = d = Vox t = 2.8 x 0.61 = 1.71 m

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