Conservation of energy and rotational energy -

[lightonahill7]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[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]

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 )

 

[rwisz]

I would approach this problem by first identifying the key principles at play: conservation of energy and rotational energy. Conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the ball's potential energy at the top of the hill is converted into kinetic energy as it rolls down the hill.

Next, I would use the equation for rotational kinetic energy, KE = 1/2 * I * ω^2, to find the ball's rotational energy. Since the ball is rolling without slipping, its angular velocity, ω, can be calculated using the equation v = ω * r, where v is the linear velocity and r is the radius of the ball.

Then, I would use conservation of energy to equate the potential energy at the top of the hill to the sum of the ball's kinetic energy and rotational energy. This would allow me to solve for the ball's velocity at the top of the hill.

Once I have the velocity, I would use projectile motion equations to find the distance the ball travels before it hits the ground. This would involve breaking the ball's motion into horizontal and vertical components and using equations for displacement, velocity, and acceleration.

Finally, I would use a table or other reference to find the moment of inertia of the ball. This would allow me to calculate the rotational kinetic energy more accurately and thus refine my calculation for the ball's velocity at the top of the cliff.

In conclusion, by applying the principles of conservation of energy and rotational energy, as well as using projectile motion equations and referencing a table for the moment of inertia, I would be able to accurately determine the distance the ball travels and its velocity before it lands at the foot of the cliff. This approach showcases the importance and usefulness of these fundamental principles in solving real-world problems in physics.