A solid uniform sphere rolls without slipping up a hill beginning at its base wi

Question

A solid uniform sphere rolls without slipping up a hill beginning at its base with 26.0m/s. The hill rises and then levels off before ending in a vertical cliff. It rises a total of 27.0m. The ball goes up the hill and then launches horizontally off cliff and landing at the original height at the hill's base. The ball's radius is 5.00cm. How fast is the ball traveling at the top of the hill? How far from the base does it land? How fast is it traveling when it hits the ground? What is the difference in angular speed between angular speed at the beginning of the problem at the base of the hill and the speed at the bottom of the fall?

Explanation / Answer

1)

initial energy Ei = kt + kr = 0.5*m*u2 + 0.5*I*w^2

I = (2/5)*m*r^2

w = v/r

kr = (1/5)*m*u^2

Ei = (7/10)m*u^2

finalenergy at the top Ef = (7/10)*m*v^2 + m*g*h

from energy conservaton

Ef = Ei

(7/10)*m*v^2 + m*g*h = (7/10)m*u^2

(7/10)*v^2 + (9.8*27) = (7/10)*26^2

v = 17.3 m/s

++++++++++++++++

along vertical

y = 0.5*g*t^2

t = sqrt(2y/g)

along horizantal

x = v*t = 17.3*sqrt(2*27/9.8) = 40.6 m

+++++++++++++++++++++++++

along vertical vy = voy + ay*t

vy = gt = 9.8*sqrt(2*27/9.8) = 23 m/s

vx = 17.3 m/s

v = sqrt(17.3^2+23^2) = 28.8 m/s

+++++++++++

wi - wf = (vi-v)/r = (26-17.3)/0.05 = 174 rad/s

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