Rolling without slipping, theory

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HaoPhysics
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Homework Statement


I am solving a question that asks for, what's the minimum coefficient of friction required for a cylinder to roll without slipping? Where the cylinder has a force F acting on its center.
Free-body diagram of a circle of radius R and mass M resting on a horizontal line, with a rightward force F at its centre and a leftward friction force f at the contact point on the bottom.

Homework Equations


Three equations: v_cm = R omega, a_cm = R alpha, and tau_net = I alpha, with subscripts cm and net.

And
Equation for friction torque: tau_fr = M g mu R, where tau_fr is the torque from friction, M is mass, g is gravitational acceleration, mu is the coefficient of friction and R is radius.


The Attempt at a Solution


From the way I understand it, the only torque acting on the cylinder is the torque from friction, since the torque from force F is acting at the center, thus resulting in 0 torque.

But the resource I'm looking at assumes that torque from friction is not the only torque acting on the cylinder. But it doesn't explain why.

Where are the other torques acting from?
 
Last edited:
Your problem statement may be incomplete. Zero is the answer as it is written now.
 
BvU said:
Your problem statement may be incomplete. Zero is the answer as it is written now.
Oops! Yeah I changed it.
 
HaoPhysics said:
But the resource I'm looking at
Tell us what you see. Our telepathic capabilities are limited.

And: what do you do with the given 'rolling without slipping' ?
 
BvU said:
Tell us what you see. Our telepathic capabilities are limited.

And: what do you do with the given 'rolling without slipping' ?

http://www.feynmanlectures.info/solutions/roll_without_slipping_sol_1.pdf
On step 6, it says, "In order for the ball not to slip, the torque on the ball from friction can not be less than the total torque on the ball when it rolls"
(Note: I tweaked the problem from a slope to a flat ground for simplification purposes, so I don't have to worry about the theta for now)

Well since it is rolling without slipping, the linear acceleration must equal to angular acceleration multiplied by radius:
Equation a_cm = Rα, where a_cm is the centre-of-mass linear acceleration, R is the radius and α is the angular acceleration.


So since the sum of torques is the torque from friction:
Handwritten-style typeset derivation of rolling friction: tau_net = tau_fr = Mg mu R; I alpha = Mg mu R; (MR^2/2)(a_cm/R) = Mg mu R; mu = g a_cm / 2.


Is this correct then?

(Sorry about the really big pictures)
 
Attachments
  • Handwritten-style derivation of rolling friction: tau_net = tau_fr = Mg mu R; I alpha = Mg mu R; (MR^2/2)(a_cm/R) = Mg mu R; mu = g a_cm / 2.
    upload_2017-5-6_19-42-46.png
    5.1 KB · Views: 476
HaoPhysics said:
Is this correct then?
Retry that last step.
Also, your answer should express μmin in terms of F, M, g and R (or any subset of those).
 
haruspex said:
That's better. But see the edit I made to my previous post.
Hm..

The only substitution which I can make is this:
Two equations side by side: mu = a_cm / 2g above, and mu = F / 2gM below, with mu on the left of each equals sign.


I am not sure how to get R in there.
It seems that the coefficient of fric. is not dependent on R.

Interesting, this follows:

Equation rendered as typeset math: F_fr = Mg mu = Mg (F / (2gM)) = F/2.
 
HaoPhysics said:
No, acm is not F/M.
HaoPhysics said:
I am not sure how to get R in there.
As I wrote, a subset of those variables is fine. It just must not involve any variables outside of the given set.
 
haruspex said:
No, acm is not F/M.

As I wrote, a subset of those variables is fine. It just must not involve any variables outside of the given set.
Oops! acm is net force / m,

Handwritten-style derivation of friction coefficient mu: mu = a_cm/2g, mu = F_net/2mg, mu = (F - f)/2mg, mu = F/2mg - f/2mg, mu = F/2mg - mu mg/2mg, mu = F/2mg - mu/2, 3mu/2 = F/2mg, mu = F/3mg.
 
Attachments
  • Handwritten sequence of equations for friction coefficient mu: mu = a_cm/2g, mu = F_net/2mg, mu = (F - f)/2mg, mu = F/2mg - f/2mg, mu = F/2mg - mu m/2mg, mu = F/2mg - mu/2, 3mu/2 = F/2mg, and mu = F/3mg.
    upload_2017-5-7_11-29-9.png
    8.9 KB · Views: 500