2011 Physics Olympiad Problem 25

  • #1
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Homework Statement


A hollow cylinder with a very thin wall (like a toilet paper tube) and a block are placed at rest at the top of a plane with inclination θ above the horizontal. The cylinder rolls down the plane without slipping and the block slides down the plane; it is found that both objects reach the bottom of the plane simultaneously. What is the coefficient of kinetic friction between the block and the plane?

Homework Equations



α =a/r
Iα = τ


The Attempt at a Solution


So for this problem, I figured that their accelerations had to be equal. The blog's acceleration was simply

gsinθ - μgcosθ

However, I'm having trouble finding the cylinder's acceleration.
Iα = τ
Summing torque around the point of contact of the cylinder (to eliminate normal force as producing a torque) I have that

τ = mgr sin θ = Iα
However, I'm not sure how to get the moment of inertia. As it's not around the Center of mass, it's not going to be MR2 for this particular reference point. Could anyone help me out here? How do I solve this problem using the COM as a reference point?
Thanks,
Thundagere
 

Answers and Replies

  • #2
ehild
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Do you know the parallel axis theorem? With respect to an axis parallel with the one going through the COM at distance d from it, the moment of inertia is I = I(COM) +md2.

But you can solve the problem also considering the rolling motion as translation of the COM and rotation about the COM. The translation is driven by the opposite forces mgsinθ and the static friction Fs. For the the angular acceleration of the rotational motion you can write the equation Iα=Fsr. You also know that a=αr in case of rolling. Cancel Fs.



ehild
 
  • #3
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So with the parallel axis theorem it would be 0.5mr2+mr2 = 1.5 mr2?
I initially took the moment of inertia about the center of mass to be mr2, but it would actually be 0.5mr2, correct? As the outer radius essentially goes to 0.
I managed to do it this time, and I think that my essential mistake was not knowing that the moment of inertia around a toilet paper cylinder was .5mr2. I assumed it would be mr2. Thanks!
 
  • #4
SammyS
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So with the parallel axis theorem it would be 0.5mr2+mr2 = 1.5 mr2?
I initially took the moment of inertia about the center of mass to be mr2, but it would actually be 0.5mr2, correct? As the outer radius essentially goes to 0.
I managed to do it this time, and I think that my essential mistake was not knowing that the moment of inertia around a toilet paper cylinder was .5mr2. I assumed it would be mr2. Thanks!
The moment of inertia of a cylindrical shell, mass M and radius R, about the axis of the cylinder is I = MR2 ,

I (1/2)MR2 .

(1/2)MR2 is for a solid cylinder.
 
  • #5
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I'm clearly doing something wrong then, as I'm arriving at the correct answer with 0.5mr2. Could you explain why? I thought a hoop was given by 0.5m(a2+b2). If b goes to 0, or essentially 0, then wouldn't it simply be 0.5mr2.

I did
Ia/r = τ

Plugging in what I thought I was equal to
(0.5)(a)(m)(R2)/R = Rmgcosθ

Cancelling and rearranging
μgcosθ = a/2 = Fs

This is the equation for the motion of the cylinder down the ramp, in terms of translational motion
gsinθ - μg cosθ = a
Plugging in and solving
gsinθ = 1.5 a
a = (2/3)(g)(sinθ)
This is the equation of the block
Since the accelerations should be the same, I plugged in the acceleration of the rolling cylinder.
g sinθ - μgcosθ = (2/3)gsinθ
Cancelling and dividing
(1/3)sinθ = μ cosθ
μ = (1/3) tanθ

What did I do wrong here?

Thundagere
 
  • #6
ehild
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I'm clearly doing something wrong then, as I'm arriving at the correct answer with 0.5mr2. Could you explain why? I thought a hoop was given by 0.5m(a2+b2). If b goes to 0, or essentially 0, then wouldn't it simply be 0.5mr2.
a is the inner radius and b is the outer one. If the wall is very thin, a≈b. So the moment of inertia of the shell with respect to its centre is I=mR2. The moment of inertia with respect to the instantaneous axis is 2mR2.
Thundagere;4204362 I did [I said:
I[/I]a/r = τ

Plugging in what I thought I was equal to
(0.5)(a)(m)(R2)/R = Rmgcosθ
The torque is Rmgsinθ with respect to the instantaneous axis.
The correct equation is 2 mR2 a/R= Rmgcosθ

ehild
 
  • #7
SammyS
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I'm clearly doing something wrong then, as I'm arriving at the correct answer with 0.5mr2. Could you explain why? I thought a hoop was given by 0.5m(a2+b2). If b goes to 0, or essentially 0, then wouldn't it simply be 0.5mr2.

Thundagere
What if b → a = R ?
 
  • #8
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Solved it this using conservation of energy! Sorry, this is an old thread, but in case anyone in the future googles this USAPhO problem, here's the solution. And thundagere, μ = (1/2) tanθ, not (1/3) tanθ.

M = mass of cylinder, m = block, R = radius cylinder, h = height on ramp, h/sinθ = distance down incline

The reasoning is that if they reach the bottom simultaneously, the velocities at the bottom must be equal, and energy is always conserved.

Mgh = Iω2/2 + Mv2/2 --> = MR2(v/R)2/2 + Mv2/2

Eventually you cancel stuff out and get this result: v2 = gh

Now for the block. Potential energy equals kinetic energy plus work done by friction

mgh = mv2/2 + μmgcosθ(h/sinθ)

gh = v2/2 + μghcotθ

Replacing v2 with gh and working towards isolating μ

gh / 2 = μghcotθ

1/2 = μcotθ

μ = (1/2)tanθ
 
Last edited:

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