A solid sphere is released from the top of a ramp that is at a height h1 = 2.50

Question

A solid sphere is released from the top of a ramp that is at a height h1 = 2.50 m. It rolls down the ramp without slipping. The bottom of the ramp is at a height of h2 = 2.19 m above the floor and the edge of the ramp has a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.160 m.

http://www.webassign.net/webassigncalcphys1/10-p-071.gif

(a) Through what horizontal distance d does the ball travel before landing?
(b) How many revolutions does the ball make during its fall?

Explanation / Answer

let v is the linear speed of the ball when it leaves the clif.

Vertical height travelled before paasing leaving the clif,

h = h1-h2

= 2.5 - 2.19

= 0.31 m

Apply conservation of energy

final kinetic energy = initial potentail energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/5)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + 0.2*m*(r^2*w^2) = m*g*h

(1/2)*m*v^2 + 0.2*m*v^2 = m*g*h

0.7*m*v^2 = m*g*h

v = sqrt(g*h/0.7)

= sqrt(9.8*0.31/0.7)

= 2.08 m/s

let t is the time taken to land from clif.

Apply, -h2 = voy*t - 0.5*g*t^2

-h2 = -0.5*g*t^2

==> t = sqrt(2*h2/g)

= sqrt(2*2.19/9.8)

= 0.6685 s

a) d = v*t

= 2.08*0.6685

= 1.39 m

b) angular speed of sphere when it leaves the clif, w = v/r

= 2.08/(0.16/2)

= 26 rad/s

no of revolutions made, N = w*t/(2*pi)

= 26*0.6685/(2*pi)

= 2.77 revoluions

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