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Friction, Kinetic Energy, & Momentum of a Sphere down an inclined plane

Posted on 2012-04-08
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Last Modified: 2016-06-25
1. The problem statement, all variables and given/known data

A solid sphere of mass 3.00 kg and radius 12.5cm rolls without slipping down an incline of
angle 13.5 degree for 250 m. Find the minimum coefficient of static friction required for a
rolling without slipping. What is the velocity of the center of the sphere at the bottom of the
incline? What is the angular momentum at that point? That is the kinetic energy at this point? Make a drawing, show the forces and torques. Indicate the torque which you are using for your calculations. Derive your formulas.

[Figures can all be rounded to 3 sig figs.]

M = 3.00 kg
R = 12.5 cm = 0.123 m
theta = 13.5
d = 250 m
h = d*sin(theta) = 58.4 m

2. Relevant equations

V = volume = 4/3(pi)R3; dV/dR = 4(pi)R2
(Rho) = M/V
Icm = (integral)r2dm
dm = (Rho)*dV
ME = Krot + Kcm - U
L=Ia(omega); Ia = Icm + Md^2 where d is the distance between the two axes

3. The attempt at a solution

Progress on problem
I found the velocity fine, and the answer matched with the solutions, but I can't seem to get  the angular momentum, coefficient of friction, or the kinetic energy at the bottom.

For friction...
I know that fs = µs*N & N = Mgcos(theta) but how do I find out what fs is?

For angular momentum...
I started with the L equation above, then subbed in the values from my previously derived work (see image) but the answer I got was way off from the correct answer.

For kinetic energy at the bottom...
It would just be 7/10*M*v2 minus the work done by friction correct? So if I can somehow find the work done by friction over the distance traveled, then I find the amount of kinetic energy left at the bottom... would that be right?

I know there are a lot of questions here, but since they're all related to the same problem, I thought I would take a shot and just put all the thoughts in my head about this problem and just have the community pick at what they feel they want to attack first, and maybe help me make a game plan for this sort of problem. Thanks in advance for any help given!



Official Answers:
V=28.6 m/s
L=4.39 kg m2/s
µs=0.0686
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Question by:Zenoture
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Accepted Solution

by:
d-glitch earned 250 total points
ID: 37823420
The sphere rolls without slipping, so the kinetic energy at every point is the sum of center-of-mass kinetic energy and the rotational kinetic energy.
All the work done by friction goes into rotational kinetic energy.  No sliding implies no heat loss.

Note that the ramp is long and shallow.  Not at all like your drawing.
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Assisted Solution

by:aburr
aburr earned 250 total points
ID: 37823493
d-glitch has asked a number of good questions which will help you to get the right answers.
Here is another related.
Just what is the amount of work done by friction? In this case the answer can be stated immediately which no calculations.
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Expert Comment

by:aburr
ID: 41668220
Both answers gave a hint of the type permitted by he homework rules and should be awarded points
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LVL 27

Expert Comment

by:aburr
ID: 41668224
Both answers gave help as permitted by the homework rules
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Expert Comment

by:aburr
ID: 41668225
See previous comment
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