Circular motion: could this happen?

  • Thread starter jryan422
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circular motion: could this happen?!?

1. Problem, given data

I have a car traveling a curve initially at 75 mi/hr. It comes to rest before it travels 350 ft along the curve. The radius of the curve is 650 ft.

I am asked if this situation is possible.

2. Equations

a = v^2/R


3. Attempt at a solution

a = v^2/R
v^2 = (a)(R)
= [(75mi/hr - 0mi/hr)(1m/s//2.24mi/hr)] [650ft(1 m / 3.28ft)]
when I find V, it's not the right answer.

This is not the correct answer. Am I setting this up correctly? Am i supposed to use s=r(angle) to determine the angle and transform this into a banking problem?
 

Answers and Replies

  • #2
Astronuc
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[(75mi/hr - 0mi/hr)(1m/s//2.24mi/hr)] [650ft(1 m / 3.28ft)]
is not correct since one is multiplying velocity (of the difference) by distance.

It does not appear that banking is to part of the problem.

Use the initial conditions to determine the coefficient of friction for the tires. The car has to stay on the curve at maximum centrifugal force which much equal the friction force. Both are proportional to the mass of the car.

Then use that to determine the maximum permissible deceleration.

Can the car decelerate from 75 mph to 0 in 350 ft? What is the necessary deceleration?

It is not necessary to convert to metric, but if one does, it's best to convert all the known values first rather than in the equation.
 
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  • #3
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wait im sorry, but i still get a wrong answer.

the normal force is equal to the gravitational force. the force of static friction is equal to the force of drag. I can only find the normal force, which is just (9.8N/kg)M (mass is not given). what forces are responsible for the inward acceleration to the center of the circle? the Force of static friction is related to the normal force, but it's impossible to determine the coefficient of SF.
 
  • #4
Astronuc
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In order for the car to stay on the curve the friction supplies the centripetal force which must equal the centrifugal force, i.e.

[itex]\mu[/itex]mg = mv2/r, but since m is the same, this simply becomes

[itex]\mu[/itex]g = v2/r, or

[itex]\mu[/itex] = v2/gr and if this is greater than 1 the car can't possibly stay on the curve.

75 mph = 110 ft/s and g = 32.2 ft/s2 and r = 650 ft.
 

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