^{2}/r

^{2}/r

^{2}/r

^{2}/r

^{2}cancel from the equation and we are left with 1/4.

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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John M. | John - Algebra TutorJohn - Algebra Tutor

Marked as Best Answer

The friction must counteract (and equal) the outward acceleration defined by:

eq 1) a = v^{2}/r

And the friction force can be found with

F= ma, subsituting eq 1) F=mv^{2}/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 mv^{2}/r

---- = -------

F2 m(2v)^{2}/r

the m, r and v^{2} 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.

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

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.

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