Centripital force and acceleration

  • Thread starter abohn1
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  • #1
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A sign on a National Forest road says: CAUTION! DANGEROUS CURVE AHEADóREDUCE
SPEED TO 20 MPH. One day, a particularly bright physics student gauged the coefficient
of friction between her tires and the road to be 0.40. Calling on her extensive knowledge
of physics and her skepticism of truth in advertising, she figured that the sign was there
to scare, not to protect, and that she could easily take the turn at 30 mph, a full 10 mph
above the posted speed limit. The turn was a tight one, 50 meters in radius.

A. Was the driver able to negotiate the turn? Why or why not?

B. How far (in mph) below the true speed limit was the posted speed limit?

C. If you take away friction, the turn would have to be banked. Find the necessary
angle for this banking, in order for our driver to safely negotiate the turn at 14
m/s.

Okay, so I'm having trouble realizing what information needs to be used and how. I found the centripetal acceleration to be 3.6 m/s^2...but without the mass of the car I'm drawing a blank as to how I can solve some of these. A point in the right direction would be great. Thanks!
 

Answers and Replies

  • #2
tiny-tim
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Hi abohn1! :smile:

(try using the X2 tag just above the Reply box :wink:)
… I found the centripetal acceleration to be 3.6 m/s^2...but without the mass of the car I'm drawing a blank as to how I can solve some of these.

Just call the mass "m".

You'll find that all the forces (including the friction force) include a factor of m, so in the end it cancels out. :smile:
 
  • #3
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Okay...thanks, but I'm still a little confused as to what equation(s) to use?
 
  • #4
tiny-tim
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Show us what you think. :smile:

(the aim is an equation for the maximum speed on a horizontal surface for a circular motion of radius 50m with coefficient of friction 0.40)
 
  • #5
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Well, I was thinking to set F=ma equal to the centripetal force F = m (v^2) / r which would work for the first 2 parts I suppose, just not sure where to put the frictional force in. Should it go to the original force equation? Or both? It seems to me like it should be relavant in both, but as you said they would then cancel out and then it would be pointless to include it in the problem.
 
  • #6
tiny-tim
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Well, I was thinking to set F=ma equal to the centripetal force F = m (v^2) / r which would work for the first 2 parts I suppose, just not sure where to put the frictional force in. Should it go to the original force equation? Or both? It seems to me like it should be relavant in both, but as you said they would then cancel out and then it would be pointless to include it in the problem.

(no, i said the mass, m, cancels out)

erm :redface: … there's no such thing as centripetal force …

your v2/r is for acceleration, which goes on the RHS of F = ma …

friction is the only force here (and it goes on the LHS).

Try again. :smile:
 
  • #7
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oh, whoops, acceleration is what i meant. soo, let's see.

F = ma
a = v^2 / r
so F = m (v^2) /r

this must be less than or equal to the frictional force...?

and the frictional force f = (coefficient) * mg
so..
m(v^2)/r = mg * (.4)
(v^2) / r = .4(9.81)
r = 50
solve for v, and v = 14.007 m/s which is about 31.33 mph so the car would stay on, and it was 11.33 mph under the actual maximum speed limit.

I found an equation for the critical velocity such that v = SQRT(rg tan(theta)) where r = 50, g=9.81, v = 14 m/s, and so I find theta to be 21.78 degrees.
 
  • #8
tiny-tim
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(i haven't checked the numbers, but …)

Yes, that looks fine. :smile:
 
  • #9
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Okay, thanks!
 

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