A solid sphere is released from the top of a ramp that is at a height h 1 = 2.30

Question

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

h1 = 2.30 m.

It goes down the ramp, the bottom of which is at a height of

h2 = 1.77 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?
m

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

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It goes down the ramp, the bottom of which is at a height of h2 = 1.77 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? m (b) How many revolutions does the ball make during its fall? rev

Explanation / Answer

A solid sphere with a diameter d = 0.14 m

       radius r = 0.07 m

vertical height h_1 = 2.3 m

ball leaves the bottom of the ramp is h_2 = 1.77 m

a ) ball travell before landing is

          h_2   = 1/2 g t^2

        1.77m =   1/2 * 9.8 t^2

         t = 0.601 s

K = 1/2 mv^2 + 1/2 I?^2

I = 2/5 mr^2

= 1/2 m v^2 + 1/2 ( 2/5 ) mr^2 ?^2

= 1/2 mv^2 + 1/ 5 m v^2         ( r = v / ? )

= 7 /10 m v^2

Potential energy = kinetic energy

m g h = 7 /10 m v^2

9.8 * (2.3 - 1.77) m   = 7 /10 * ( v^2 )

   v = 2.724 m/s

distance d = v * t

                     = 2.724 m/s * 0.601 s

                   = 1.637 m

b ) ? = ? t

           = v t / r

           = d / r

          = 1.637 m / 0.07 m

         = 23.39 radians

= 23.39/2*pi rev

= 3.72 rev

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