A solid uniform rolls without slipping up a hill beginning at its base with 26.0

Question

A solid uniform 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.

A) How fast is the ball traveling at the top of the hill? 17.25 m/s

B) How far from the base does it land? 40.47m

C) How fast is it traveling when it hits the ground? 28.76 m/s

D)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

Part(D)
Rolling w/o slipping means, Angular Speed = V/R

Where V is linear Velocity
R is Radius = 5.0*10^-2 m

At the Beginning -
Angular speed = 26.0/(5.0*10^-2)
Angular speed, i = 520 rad/s


At the bottom of the fall -
Angular speed = 28.76/(5.0*10^-2)
Angular speed f = 575.2 rad/s

Difference in angular speed = f - i
Difference in angular speed = 575.2 - 520 rad/s
Difference in angular speed = 55.2 rad/s

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