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

[Zenoture]

Homework Statement

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
θ = 13.5
d = 250 m
h = d*sinθ = 58.4 m

Homework Equations

V = volume = 4/3πR³; dV/dR = 4πR²
ρ = M/V
I_cm = ∫r²dm
dm = ρ*dV
ME = K_rot + K_cm - U
L=I_aω; I_a = I_cm + Md^2 where d is the distance between the two axes

The Attempt at a Solution

http://dl.dropbox.com/u/22657586/phys130_mt2_f10_num2.JPG

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 f_s ≤ μ_s*N & N = Mgcosθ but how do I find out what f_s 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:

Spoiler

V=28.6 m/s
L=4.39 kg m²/s
μ_s=0.0686

 

[collinsmark]

Hello Zenoture,

Welcome to Physics Forums!

Homework Statement

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
θ = 13.5
d = 250 m
h = d*sinθ = 58.4 m

Homework Equations

V = volume = 4/3πR³; dV/dR = 4πR²
ρ = M/V
I_cm = ∫r²dm
dm = ρ*dV
ME = K_rot + K_cm - U
L=I_aω; I_a = I_cm + Md^2 where d is the distance between the two axes

The Attempt at a Solution

http://dl.dropbox.com/u/22657586/phys130_mt2_f10_num2.JPG

I found the velocity fine, and the answer matched with the solutions,

Okay, so far so good.

For friction...
I know that f_s ≤ μ_s*N & N = Mgcosθ but how do I find out what f_s is?

You've found the velocity of the ball already. You can use your kinematics equations (for uniform acceleration) to find the ball's acceleration.

The ball is accelerating down the ramp. So the sum of all forces are not equal to zero. The sum of all forces are equal to the ball's mass times its acceleration: Newton's second law. Use your free body diagram for this. What are the forces acting on the ball?

Alternatively, if you wanted to, you could calculate the ball's angular acceleration in the same way (using, in part, the ball's initial and final angular velocity), and then use Newton's second law equivalent in terms of torques. What torques are acting on the ball? Which forces cause torque on the ball?

I suggest working it both ways to double check your answer.

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.

Um, the formula you used for moment of inertia on this part was

I_a = I_cm + Md².

Well, okay. It's not the formula I would have started with, but let's go with it anyway. That formula is for some sort of massive shape (could be a ball) pivoting on a point that is not necessarily the center of mass. For example, it could be a ball attached to one end of a massless rod, and the whole thing pivots at the other end of the rod.

So let's apply it to this problem. What is the pivot point of the ball? Is it not already rotating around its center of mass? If it is already rotating around the center of mass, what does that tell you about d in the formula?

You've already calculated the ball's moment of inertia, I_cm = (2/5)Mr².

Later in your handwritten work, you seem to be implying that ω = R/v. That's not right. ω = v/R. Just remember that the angular velocity is proportional to the linear velocity.

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?

Friction doesn't do any work, unless the ball start's sliding.

If the ball is not sliding, then the part of the ball that is touching the ramp is stationary with respect to the ramp (that's kind of the definition of not sliding). So the distance of the slide is zero, meaning the work done by friction is zero.

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:

Spoiler

V=28.6 m/s
L=4.39 kg m²/s
μ_s=0.0686