Sphere rolling up an incline then back down

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Homework Help Overview

The discussion revolves around a sphere rolling up an incline after traveling along a horizontal surface. The problem involves analyzing the motion of the sphere, including its maximum distance on the incline, the time taken to roll back down, and its final velocity upon returning to the horizontal surface. The subject area includes concepts from mechanics, specifically energy conservation and rotational motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using energy equations and conservation of energy to analyze the motion of the sphere. There are attempts to derive expressions for kinetic and potential energy, and questions arise regarding the correctness of these equations. Some participants explore the relationship between linear and angular motion, while others question the implications of using initial and final velocities in their calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's equations and reasoning. Some guidance has been offered regarding the expressions for energy and the relationships between variables. There is a recognition of the need to clarify the use of initial and final velocities in the context of the problem.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for solving the questions. There is an emphasis on deriving relationships based on given variables, and some assumptions about the nature of the motion are being questioned.

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A sphere or mass, m and radius r rolls along a horizontal surface with a constant velocity, Vi approaches an incline with (angle theta). ie bottom angle :smile: If it rolls without slipping,
a) what is the maximum distance,x it will travel on the incline?
b) If it begins to roll back down, find the time it takes to get to horizontal surface.
c)What will be its final velocity. ie. at time it gets to horizontal surface.

My solution:

I want to write energy equations and use conservation of energy to solve it.

K.E. = 1/2M(Vi^2) + 1/2I(w^2) where I = 2/5 r^2
P.E = mgh(max) - kNx where k is coeficient of friction and N is normal force

using trig, h(max) = xsin(theta) so x = h(max)/sin(theta)
so
P.E = mgh (max) - [kNh(max)] /sin(theta) ; N = mgcos(theta)
P.E = mgh(max) - [kmgcos(theta)h(max)]/ sin(theta)

h(max) is the vertical distance traveled ie. less than h and
x is distance traveled on incline i.e less than d

Are my equations right? and if so,the way to go now is to substitute w= v/r, set K.E = P.E and solve for x?
 
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shizzle said:
I want to write energy equations and use conservation of energy to solve it.
Good. That's how to solve part a).

K.E. = 1/2M(Vi^2) + 1/2I(w^2) where I = 2/5 r^2
I = 2/5 M r^2
P.E = mgh(max) - kNx where k is coeficient of friction and N is normal force
The friction does no work, since it rolls without slipping.

using trig, h(max) = xsin(theta) so x = h(max)/sin(theta)
Right.
Are my equations right?
Correct your expression for PE and you're good to go.
and if so,the way to go now is to substitute w= v/r, set K.E = P.E and solve for x?
Right.
 
Hey Doc Al,
Thanks. I think i figured out the 3rd part.

I first found the acceleration.
since w = v/r ; v = wr
a = w(dot -on top of it--hehe) r
w(dot) = a / r

I w(dot) = Fr where F is friction
ma = mg sin (theta) -F

If you plug expression for F into it
a = 5/7g sin(theta)

Now, knowing acceleration, i just separate variables and integrate to find velocity

and i got it to be 5/7gsin(theta) t

If this is right, i'll need to figure out the time it takes to get down. do i just make t the subject? that seems trivial...any help?
 
shizzle said:
If you plug expression for F into it
a = 5/7g sin(theta)
Looks good to me.

Now, knowing acceleration, i just separate variables and integrate to find velocity

and i got it to be 5/7gsin(theta) t
Right.

If this is right, i'll need to figure out the time it takes to get down. do i just make t the subject? that seems trivial...any help?
It's as easy as you think it is. :smile: [itex]V = 5/7 g sin\theta t[/itex], since you know the final speed [itex]V = V_i[/itex], just solve for t.
 
okay, but when i solve for t, my expression will involve v not vi. Should i then make vi the subject of my x expression from a) and plug vi into my t? (the vi expression will involve v meaning my final t expression will involve v)

I'm also not really sure why you say v = vi at the end. Is it because the ball goes back to rolling on the horizontal surface? Thanks. I'm almost there:smile:
 
shizzle said:
I'm also not really sure why you say v = vi at the end. Is it because the ball goes back to rolling on the horizontal surface?
Yes. Since the ball started up the incline with a linear speed of Vi, that's what it will end up with when it rolls back down.
 
so t = 7v/5gsin (theta) but v = vi so
t = 7vi/5sin(theta)?

It just seems weird that we're suddenly replacing the velocity with which it hits the ground with its initial velocity (even though i understand that vf = vi)

Is there something I'm missing? or am i just thinking too hard?
 
shizzle said:
so t = 7v/5gsin (theta) but v = vi so
t = 7vi/5sin(theta)?
Right (but don't leave out the g).

It just seems weird that we're suddenly replacing the velocity with which it hits the ground with its initial velocity (even though i understand that vf = vi)

Is there something I'm missing? or am i just thinking too hard?
It may seem weird, but vi is the only information you are given, so your answer had better be in terms of it!

If you want to, why not figure the time it takes to go down the incline using the distance x that you already calculated? (You'd better get the same answer!)
 

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