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Centripetal force for off-centered cylinders rolling down a curve

  • Thread starter caesium
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  • #1
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Homework Statement:

Let's say we have a cylinder of radius r with non-uniform density, resulting in a center of mass shifted away from center of geometry. Let's assume that this center of mass mid-way between center and circumference. So, r_cm = r/2.

Now, let's say this cylinder is rolling down on a curved path of radius R + r. Assume no slip condition.

What is the centripetal force on this cylinder? Does it have "two centripetal forces", from which we find a resultant?

Relevant Equations:

There are no equations here.
My initial attempt: Total Centripetal force on the cylinder would be given by $$\textbf{F}_{net} = mR\omega^2 \textbf{e}_1+mr_{cm}\omega^2 \textbf{e}_2$$ where the vectors e_1 and e_2 have magnitude 1 and point radially outwards (and continuously changing as the cylinder rolls down) as marked in the sketch below. Please advise. Thank you.

nmjg667dcd-jtls3.png
 

Answers and Replies

  • #2
That is a strange problem. First of all, what is definition of a centripetal force acting on a rigid body ?
 
  • #3
haruspex
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I assume you mean R>r.

Is there any guarantee that the cylinder will remain in contact with the surface?

As @wrobel asks, define centripetal force; but first define centripetal acceleration.
 
  • #4
I think that there are only two forces acting on the cylinder: the reaction force from the incline and the gravity
 
  • #5
haruspex
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I think that there are only two forces acting on the cylinder: the reaction force from the incline and the gravity
Yes, but centripetal force does not have to be one or the other.
 
  • #6
I was taught long ago that forces appear between interacting material bodies
 
  • #7
haruspex
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I was taught long ago that forces appear between interacting material bodies
And the sum of such forces is also a force. As is a component of the sum.
 
  • #8
And the sum of such forces is also a force
and to which point of the cylinder is this force applied? and why?
 
  • #9
haruspex
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and to which point of the cylinder is this force applied? and why?
As I wrote, given the velocity and the trajectory, it is better to start with the definition of centripetal acceleration. Multiplying that by the mass gives the centripetal force.
 
  • #10
As I wrote, given the velocity and the trajectory, it is better to start with the definition of centripetal acceleration. Multiplying that by the mass gives the centripetal force.
Let us have an inertial coordinate frame. For the cylinder I want to write the angular momentum equation about the origin of this frame. And how should I include the torque of the centripetal force into this equation? If it is force indeed, I must put its torque into the angular momentum equation
 
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  • #11
haruspex
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By definition, centripetal acceleration is that component of linear acceleration which is orthogonal to the velocity.
In principle, we can find the velocity and acceleration of the mass centre. But there is insufficient information here since we do not know the moment of inertia.
 
  • #12
I think that notion "centripetal force" does not make sense for rigid bodies just for particles
 
  • #13
haruspex
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I think that notion "centripetal force" does not make sense for rigid bodies just for mass points
That would be somewhat heretical.
 

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