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Motion along curved path

by TraceBusta
Tags: curved, motion, path
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TraceBusta
#1
Sep17-04, 12:50 PM
P: 35
i'm stuck on this one.
some bootleggers race from the police down a road that has a sharp, level curve with a radius of 30 m. As they go around the curve, the bootleggers squirt oil on the road behind them, reducing the coefficient of static friction from 0.66 to 0.18. When taking this curve what is the maximum speed of:
a) the bootlegger's car? b) the police car?

My free body diagram has w=mg in the negative y, and Fnormal in the positive y...cancelling out but because the problem doesn't involve any masses i assume that you don't need to find the force normal or weight. I dont know what else to draw on the free body diagram other than the force of static friction pointing to the center of the curve? I dont know where to start with this.
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arildno
#2
Sep17-04, 12:55 PM
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Welcome to PF!
Try to answer the following question:
What force must provide the centripetal acceleration?

And secondly, how is centripetal acceleration related to velocity and radius of curvature?
TraceBusta
#3
Sep17-04, 01:21 PM
P: 35
the static friction provides the centripital acceleration? and i know that a=v^2/r so i need to find a somehow in order to solve for v. If i'm right with the static friction causing the centripital acceleration i still don't know how to solve for a.

arildno
#4
Sep17-04, 01:53 PM
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Motion along curved path

Use F=ma! (insert v^2/r, and use the static friction as F)
TraceBusta
#5
Sep17-04, 02:29 PM
P: 35
using F=ma I dont understand how I could use that if there is no mass given, also the static friction is just a coefficient given so how could i use that as F?
arildno
#6
Sep17-04, 02:41 PM
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Static friction F is a force, the static friction coefficient (mu) is a coefficient.
F=(mu)mg, so you get:
(mu)mg=mv^2/r
TraceBusta
#7
Sep17-04, 03:52 PM
P: 35
Thanks! I understand it now. I used r*(mu)*g=v^2 to solve for v.

I used the same setup for a following question, but I for some reason keep getting the wrong answer. The question is
In an amusement-park ride, riders stand with their backs against the wall of a spinning vertical cylinder--this ride is sometimes called a Silly Silo or a Cyclone. The floor falls away and the riders are held up by friction. If the radius of the cylinder is 4.5 m, find the minimum number of revolutions per minute necessary to prevent the riders from dropping when the coefficient of static friction between a rider and the wall is 0.46.

I use 4.5m*.46*9.81m/s^2=v^2 and v=4.506 m/s
Then I multiplied that by (60 s/1 min) to get 270.36 m/min. The circumfrence of the circle is 28.274 m from 2pi*r. So I divided that into the velocity and get 9.562 rev/minute but that is the wrong answer.
Pyrrhus
#8
Sep17-04, 04:46 PM
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Remember in the cyclone the cetripetal force will be the normal force, (forces on x-axis) and on the y-axis we got the friction force pointin up and the weight pointing down.

Also a hint [tex] F_{f} = \mu N [/tex]
TraceBusta
#9
Sep17-04, 07:03 PM
P: 35
thanks, got it. I solved for v in (mu)(v^2/r)=g then found the rev/minute with that.
austin zitro
#10
Sep19-04, 10:46 PM
P: 5
How did you get mu(v^2/r)=g? It seems like it should be Ff=ma which would be mu*m*g=m(v^2/r). Btw, are you in physics 152 at Purdue?
Pyrrhus
#11
Sep19-04, 11:26 PM
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Quote Quote by austin zitro
How did you get mu(v^2/r)=g? It seems like it should be Ff=ma which would be mu*m*g=m(v^2/r). Btw, are you in physics 152 at Purdue?
Read my explanation.


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