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What must be the friction?

Harry Hotrod rounds a corner in his sports car at 50 km/h. The friction force holds him on the road. If he has twice the speed, what must be the friction force to prevent him from skidding off the road?

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The friction is proportional to the square of the speed. If the speed is doubled, then the friction force is quadripled. No other calculation is needed.

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2 Answers

The friction must counteract (and equal) the outward acceleration defined by:
 
eq 1)    a = v2/r
 
And the friction force can be found with
 
F= ma, subsituting eq 1)    F=mv2/r
 
For this problem mass and radius are constants. 
 
Let F1 equal the friction force at 50 km/h.
Let F2 equal the friction force at 100 km/h  (2x50)
 
If we set this up as a ratio:
 
F1        mv2/r
---- =  -------
F2        m(2v)2/r
 
the m, r and v2 cancel from the equation and we are left with 1/4. 
 
The friction force at 100 km/h must be 4 times larger than it was at 50 km/h.

Comments

If I were answering this I would be careful about saying there was "outward acceleration".
Part of the issue with circular motion of any kind is breaking a student's pre-conception and getting a student to see and believe that there are no forces or accelerations in the outward direction.  Conceptually they need to believe there it is only a centripetal force that causes a centripetal acceleration and circular motion.
Robert you are absolutely correct. The Physics behind why a passenger  

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Let f be the friction force. Apply Newton's second law towards to the center of the rotation.
f = mv^2/r, where f is the only force in centrilpetal direction.
If you double v, then f is quadripled.