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12-20-2009   #1 (permalink)
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Default [Physics] Centripetal Force Problem

I need help with this problem:

A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static frictional force by a factor of three. If the car is to continue safely around the curve, to what speed must the driver slow the car?

Formula: Centripetal Force = mass X velocity squared / radius
 
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12-21-2009   #2 (permalink)
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There is really only one force acting on the car: A centripetal force mv^2/r, which is pointing towards the center of the turn. This centripetal force is due to the effect of friction.
In other words, as the car goes around the turn, there is an effective centripetal force pointing towards the inside of the turn, and this force is due to the friction between the car's tires and the road.

Since these forces are one and the same, you'll want to set the equation for the car's centripetal force and frictional forces equal:

M V^2 / R = mu[k] * N

Normal force = M*g
yielding

M * V^2 / R = mu[k] * M * g

Doing this allows us to solve for mu[k].

mu[k] = V^2 * [ 1 / (g*r) ]

g is a constant on the planet earth, as is r (the car is still going around the same curve). Thus mu[k] is being related to the velocity.

if the frictional force is decreased by 1/3, then the velocity must be adjusted accordingly, by a factor of sqrt (1/3).
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12-21-2009   #3 (permalink)
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I'm kinda confused about the static/kinetic friction thing...It's moving, yet it's being acted upon by static friction.

I set it up like this though

[i] = subscript i (initial)
[f] = subscript f (final)

F[i] = mv[i]^2/r
F[f] = mv[f]^2/r
F[i]/F[f]=3
(mv[i]^2/r)/(mv[f]^2/r) = 3
v[i]^2/v[f]^2 = 3
v[i]/v[f] = sqrt(3)
v[f] = v[i]/sqrt(3)

Basically same as above but you don't really need to use the formulae of friction, meaning it works for other forces too that provide the centripetal acceleration. As long as you know the ratio between the two forces, you know the ratio between the two speeds.
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12-21-2009   #4 (permalink)
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static friction is always present in motion problems involving wheels. IF you look at the point of contact between the wheel and the ground, there is no "sliding"; each point of the wheel is stationary, while the rotation of the wheel provides the forward motion. Sliding = kinetic friction. Since each tire point only hits one point on the ground, then the only friction present is static friction.
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12-30-2009   #5 (permalink)
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Be sure to draw a free body diagram before you attempt these kind of problems.

And on a more humorous note: xkcd - A webcomic of romance, sarcasm, math, and language - By Randall Munroe
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