# Conservation of energy and rotational energy - please help!

#### lightonahill7

1. The problem statement, all variables and given/known data

A solid, uniform spherical ball rolls without slipping up a hill. At the top of the hill it is moving horizontally, and then it goes over a vertical cliff. How far from the foot of the cliff does the ball land, and how fast is it moving before it lands.

Use conservation of energy to find the velocity of the ball at the top of the cliff. Then use projectile motion to find how far it falls. Use table to find moment of inertia of the ball.

2. Relevant equations

3. The attempt at a solution

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#### Delzac

Welcome to PF lightonahill7!

What are your initial thoughts so far? We will guide you from where you are facing problem.

Delzac

#### lightonahill7

Vo = 25 m/s
Please look over my solution to see if I am right.

h = 28 m
I = 2/5mr^2
w (omega) = v/r initital
wfinal = vfinal/r

Conservation of energy

1/2mvo^2 + 1/2Iw1^2 = mgh + 1/2mvfinal^2 + 1/2Iwfinal^2

This is equation one.

2 unknowns w (omega) and Vfinal

Rolling w/o slipping means that w= v/r

1/2Iw^2 = 1/2(2/5mr2)(v/r)2 = 1/5mv^2 - Equation 2

Substitue into equation 1 and solve for vfinal

1/2mvo^2 + 1/5mvo^2 = mgh + 1/2mv2^2 = 1/5mv2^2

7/10mv1^2 = mgh + 7/10mv2^2

v2^2 = (v1^2 - 10/7gh)^1/2

= [(25}^2 - 10/7(9.8)(28)]^1/2

= (233)^1/2

v2 = 15.26 m/s

Determine the time in the air?

y-component

Voy = 0

ay = 9.8 m/s^2

y-yo = 28 m/s

28 = voy + 1/2 ayt^2

t^2 = 28/4.9

t^2 = 5.71

t=2.39 s

x-component - distance from cliff

d = vfinal*t
= 15.26 m/s(2.39 s)
= 36.47 m

How fast is it moving before it lands?

vy = voy + ayt
= (9.8)(2.39)
= 23.42 m/s

vx = Vox = 15.3 s

v = [(23.4)^2 + (15.3)^2]
= (781.65)^1/2
= 27.96 m/s

#### tiny-tim

Science Advisor
Homework Helper
hi lightonahill7!

(have an omega: ω and try using the X2 and X2 icons just above the Reply box )

yes that's all ok (though you should probably have less sig figs in the answers )

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