Finding Tension in Horizontal circular motion with friction

AI Thread Summary
The discussion revolves around calculating the minimum strength of a cable connecting a car to a rotating pole while considering friction. The car's maximum speed is given, and calculations indicate that it will skid without additional centripetal force, which is provided by both static friction and cable tension. The user is confused about the role of kinetic friction, which only becomes relevant once the car begins to skid. It is clarified that the tension in the cable must be sufficient to prevent skidding; otherwise, it will fail. Understanding these forces is crucial for determining the cable's minimum strength.
applejuicesam
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Homework Statement


A car of mass (m) is attached to a cable of length (L) to a rotating pole. The car will speed up and travel in a circle. If the maximum speed of the car is vmax, what should the minimum strength of the cable connecting the car to the pole be? Note the surface is not frictionless.m = 137.7 kg, L = 5.5 m, μK(road-tires) = 1.13, μS(road-tires) = 1.70, vmax = 10.7m/s

** I'm confused because the car is attached by a cable to the pole, but I also have values for static and kinetic friction. If the car is attached to the pole, does friction even matter? Since Ff=Fc and Tension=Fc, doesn't Ff=Tension? And where does kinetic friction factor into this?

Homework Equations


Fc=mv2/r, Fc=Ff, Fc=Tension, ac=v2/r

The Attempt at a Solution


My FBD attempt. (Still confused by the friction).
http://imgur.com/asxof5h [/B]
 
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Hints
On a car (without a pole) what normally provides the centripetal force needed to go around a corner?
What limits how fast the car can go around the corner?
If the car was to go faster than that limit it will skid because there is insufficient centripetal force - what could provide the additional centripetal force?
 
CWatters said:
Hints
On a car (without a pole) what normally provides the centripetal force needed to go around a corner?
What limits how fast the car can go around the corner?
If the car was to go faster than that limit it will skid because there is insufficient centripetal force - what could provide the additional centripetal force?
-The force of friction allows it to go around a corner.
-The value of static friction limits the speed of the car. (Higher coefficient of static friction allows for the turn to be navigated at greater speed).
-If the car exceeds the max speed allowed by the force of friction, it starts to skid and the force is kinetic friction...and kinetic friction points opposite to the direction of motion.

*I'm still not sure how the cable tension is involved in this?
 
It's hard to give another hint without giving the whole answer.

See last line of my post above.
 
What could provide the additional centripetal force to stop it skidding when going too fast for friction alone?
 
CWatters said:
It's hard to give another hint without giving the whole answer.

See last line of my post above.

CWatters said:
What could provide the additional centripetal force to stop it skidding when going too fast for friction alone?
The tension of the attached cable...but if this is so, why are you assuming that the car is skidding in the first place? How do I know that the max velocity given is enough to cause it to skid, causing the Tension of the cable to aid in centripetal force?
 
I'm not assuming it will skid. In fact it can't skid unless the rope fails.

You don't need to know exactly how fast the car is going to write an equation for the tension in the cable and hence its minimum strength. You can always add a note that below a certain speed the minimum strength of the rope is zero.
 
Actually I see they tell you the Max speed. Have you tried calculating if it will skid?
 
CWatters said:
Actually I see they tell you the Max speed. Have you tried calculating if it will skid?
Okay. I calculated that the max speed is 9.57 m/s which means that it will skid without any additional centripetal forces! So right now I'm conceptualizing that the Force of static friction and the tension in the cable add together keep the car moving at the given max speed, calling it Fc total. So I can find the tension of the cable by finding the Fc total and subtracting the Fstatic friction. If I'm reading this all correctly, does the kinetic friction factor in here at all?
 
  • #10
applejuicesam said:
Okay. I calculated that the max speed is 9.57 m/s which means that it will skid without any additional centripetal forces! So right now I'm conceptualizing that the Force of static friction and the tension in the cable add together keep the car moving at the given max speed, calling it Fc total. So I can find the tension of the cable by finding the Fc total and subtracting the Fstatic friction. If I'm reading this all correctly, does the kinetic friction factor in here at all?
That's the correct reasoning.
Re kinetic friction, that only comes into play once it starts to skid. What will happen to the centripetal force provided by friction if that happens? If the cable is not strong enough to prevent skidding, will it be strong enough to hold a car which is starting to skid?
(Note also that if we take the cable as being completely inextensible them skidding would imply the cable has already failed.)
 
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