Suppose a small coin is on a flat horizontal rotating turntable

[bullroar_86]

and does not slide off.

in this case does centripetal force = frictional force?



I think the answer is yes, but want to make sure.

thanks

 

[whozum]

centripetal force < frictional force.

edit: (centripetal force = fricitonal force) < or = MAXIMUM frictional force which is [itex] \mu_s N [/tex]

 

[futb0l]

Yeah, it's the only force that is applied to the coin, isn't it?

 

[Ouabache]

Not necessarily... How about when the turntable is rotating slower, the coin still does not slide off.. Would centripetal force = frictional force at that moment also?
What would be a better answer?

edit: looks like several other folks added their 2 cents before I sent mine. I try to help by getting you to think a bit..

 

[whozum]

Yeah, it's the only force that is applied to the coin, isn't it?

The forces on the coin would be gravity, normal, centripetal, and frictional.

 

[SpaceTiger]

As far as I can tell, friction is the only thing that can act as the centripetal force in this situation. Gravity and the normal force aren't acting in the right direction. From what I understand of static friction, it applies a force only large enough to resist the other forces involved (in this case, the force rotating the turntable). Thus, decreasing the speed of rotation will only decrease the force that static friction has to apply in order to resist it. If you continually increase the rotation rate, you'll reach a point at which static friction can no longer hold the coin and it will move outwards (slowed somewhat by kinetic friction).

 

[Ouabache]

Actually, in this example, centripetal force may be < or = frictional force.
(when these forces are equal, there is no movement of the coin)

 

[SpaceTiger]

Actually, in this example, centripetal force may be < or = frictional force.
(when these forces are equal, there is no movement of the coin)

What I'm saying is that static friction doesn't have a single value of force. It acts to balance the other forces, preventing motion. Otherwise, you would have friction "moving" objects and we know from everyday experience that this doesn't happen.

 

[whozum]

Centripetal force magnitude < or = to maximum frictional force magnitude. After that point, the object is no longer in uniform circular motion

 

[SpaceTiger]

Centripetal force magnitude < or = to maximum frictional force magnitude. After that point, the object is no longer in uniform circular motion

Note the word in bold. This is equivalent to what I said in my first and it is not consistent with the statement that the frictional force is less than the centripetal force.

 

[whozum]

Yeah I just wanted to clear up my statement. We are saying the same thing. The frictional force is the centripetal force up to a maximum, after which alternate motion applies.

The less than applies for rotation rates where the centripetal force required is less than the maximal frictional force.

 

[Doc Al]

centripetal force < frictional force.

As SpaceTiger made clear, if the coin does not slide centripetal force = frictional force. I think you meant to say that for static friction $F_f \leq \mu N$.

The forces on the coin would be gravity, normal, centripetal, and frictional.

"centripetal" is not a type of force, it is a direction. Since the coin executes uniform circular motion, the net force on it (the vector sum of weight, normal, and frictional forces) is centripetal.

 

[Curious3141]

The centripetal force on the coin is always equal to the static frictional force b/w coin and turntable while the coin remains stationary. The point at which the equality is violated is when the maximal frictional force is reached, and this is the limiting friction, given by the coeff of friction multiplied by the normal force of the turntable on the coin. Beyond this point, the coin begins to slide, and that state is maintained because kinetic friction is lower than static friction.

 

[whozum]

As SpaceTiger made clear, if the coin does not slide centripetal force = frictional force. I think you meant to say that for static friction $F_f \leq \mu N$.

Thanks I think I corrected myself already, I should try to express my ideas more carefully.

"centripetal" is not a type of force, it is a direction. Since the coin executes uniform circular motion, the net force on it (the vector sum of weight, normal, and frictional forces) is centripetal.

Good point, I didnt think of that.

 

[Ouabache]

This was some very insightful discussion.. Though perhaps bullroar_86 may have gotten a little more than he bargained for..
bullroar_86, did we answer your question?

 

[bullroar_86]

hahah.. yes I did get a little more than I asked for.

and yes my question was answered, and the problem has been finished


thanks for the help