# Spinning friction for a small point

Hey guys, this is not so much a homework question as it is a thought that i'm pondering.

## Homework Statement

If we have a dancer that is spinning on her toes (wherein she is wearing a metal shoe with a very very very sharp tip) with her CG located above that tip, what is the contact friction?

## Homework Equations

F= mu_k*N (where in this case N=mg)

## The Attempt at a Solution

I'm wondering if an infinitesimally small point will give 0 friction force?

in which case in reality, the point of rotation has a certain contact area (say 0.01mm^2 radius) with the normal force as mg/area?

Cheers for all the help ;)

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Homework Helper
Hi James!

The force of friction is the same, whatever the contact area.

However what matters here is the loss of energy, which is the work done by the friction force …

work done = force times displacement, and the narrower the area of contact, the smaller the displacement (when spinning on the spot) …

so the skater's loss of energy will be proportional to the width of the area of contact.

What do you mean by the force of friction is the same, whatever the contact area? If a car brakes and has tiny wheels, shouldn't it have less stopping force? (I can see where you're coming from with the formula, but im wondering why it is so)

tiny-tim
Homework Helper
Hi James!
What do you mean by the force of friction is the same, whatever the contact area?
It just is.

The force of friction depends only on the normal force …

that doesn't depend on the area of contact, so neither does the friction.

(For rubber, it's slightly different, for reasons I forget but which have to do with the deformation of the rubber … you can look that up! … but that probably wouldn't apply to the base of a dancer's shoe.)

Oh wow that's interesting. So effectively I can slide an object with 3 sharp points of contact and another with equal mass but 15 points of contact and it wouldn't make a difference if they are all rigid? And it doesn't matter if it's spinning or translating?

For friction in Lagrangian equations, (or Euler Lagrange,) the nonconservative force for the friction in the spinning shoe is (mu)*m*g? [no cos(theta) because of the gyroscopic forces]

tiny-tim
Homework Helper
So effectively I can slide an object with 3 sharp points of contact and another with equal mass but 15 points of contact and it wouldn't make a difference if they are all rigid? And it doesn't matter if it's spinning or translating?
Yup!

except they'd better not be sharp, or they'll dig in and change the coefficient … keep them smooth but small!
For friction in Lagrangian equations, (or Euler Lagrange,) the nonconservative force for the friction in the spinning shoe is (mu)*m*g? [no cos(theta) because of the gyroscopic forces]
dunno, pass

Hi guys

Well i have read this thing at many places --- The force of friction don't depend on contact area ----

but i still have some doubt. because i just did this with by book, on the flat base i felt that i have to apply more force to make it move. but when on the smaller, force was less!!!

And can anyone tell me the origin of static friction?

tiny-tim
Homework Helper
i just did this with by book, on the flat base i felt that i have to apply more force to make it move. but when on the smaller, force was less!!!
you're probably digging the book in at the edge

alternatively, the coefficient of friction may be different on different sides … perhaps you should use two identical books, first side-by-side, then one atop the other?
And can anyone tell me the origin of static friction?
hmm … start a new thread!

Hi cupid.callin,

My guess (from what I've been reading) is that because in the theoretical we assume that the contact is perfectly flat. However in reality, for an object with a larger surface area, there will be more microscopic pimples/dimples/ridges that lock together with the contact surface.

tiny-tim, I saw your comment before on the Euler-Lagrange equations so I thought you might be able to answer :P perhaps the question is more like: "so in formulating the equations of motion of the spinning dancer, the spinning friction force is just F=umg without any dependence on spin rate?"

tiny-tim
Homework Helper
Hi James!

(have a mu: µ )
perhaps the question is more like: "so in formulating the equations of motion of the spinning dancer, the spinning friction force is just F=umg without any dependence on spin rate?"
Yeees, the force is just µmg …

but what matters in spinning is the torque.

So in reality, friction does depend on the surface area ??!!

tiny-tim, oh dear. Physics is so mean to me :ρ now τorque comes in. Haha. Isn't torque irrelevant if it's spinning on a tip? (or perhaps just close to zero)

tiny-tim