Circular Motion,Uniform motion

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  • #1
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1. Homework Statement [

An Aston Martin V8 Vantage sports car has maximum "lateral acceleration" of 0.96"g's" ,after which the car begins to skid out of its turn.

Calculate the coefficient of static friction between the car's tyres and the road.


Homework Equations





The Attempt at a Solution



where Us is coefficient of friction

Fnet(vertical)=n-mg=o
Fnet(radial)=fs= Us*N=m(v^2/r)
Us*mg=m(v^2/r)
Us=V^2/r*g
=a(r)/g
= (0.96*9.81)m/s^2/9.81m/s^2
=0.96
 

Answers and Replies

  • #2
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It's correct.
 
  • #3
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1. Homework Statement [

An Aston Martin V8 Vantage sports car has maximum "lateral acceleration" of 0.96"g's" ,after which the car begins to skid out of its turn.

a)Calculate the coefficient of static friction between the car's tyres and the road.
b)Determine the minimum radius of unbanked curve the car can negotiate while travelling at a constant 120km/h.
c)If the road is banked at 15° determine the range of speeds at which the car can safely negotiate the curve in wet weather ,when the coefficient of static friction is reduced by 20%.


Homework Equations



v=√rgtan(θ)


The Attempt at a Solution




b) r=v^2/a(r)
=(33.3m/s)^2/(0.96*9.81)m/s^2
=117.98m
The radius of the unbanked and the banked will be equal?
No friction is needed in a banked road because the horizontal component of the normal force helps the car to turn?
c)If so,then
v=√117.98m*9.81m/s^2*tan(15°)
=17.61m/s
So how do i get the "range of speeds"?
 
  • #4
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"No friction is needed in a banked road because the horizontal component of the normal force helps the car to turn?"

Not a true statement. If centrifugal force component exceeds the sum of frictional force and component of force down the slope, the car slides.
 
  • #5
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For a range of speeds, what is the minimum speed the vehicle can travel without sliding down the slope? That is the lower limit.
 
  • #6
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For a range of speeds, what is the minimum speed the vehicle can travel without sliding down the slope? That is the lower limit.

V=√rgtanθ is only valid when the car is not sliding I think,so this "v" will be the minimum speed.But for the upper limit I am not sure.
 
  • #7
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"V=√rgtanθ is only valid when the car is not sliding I think,so this "v" will be the minimum speed.But for the upper limit I am not sure. "

The above formula is the speed when centrifugal force balances the force of the vehicle wanting to slide down the hill. It's the speed when there is no friction.

Determine the forces on the tires where they meet the pavement. You have 3 forces that you consider in order to determine the maximum speed given the banking and friction.
 
  • #8
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"No friction is needed in a banked road because the horizontal component of the normal force helps the car to turn?"

Not a true statement. If centrifugal force component exceeds the sum of frictional force and component of force down the slope, the car slides.


In the questions that I did so far on banked curves the centrifugal force is the horizontal component of the normal.And won't the frictional force be a one of centrifugal forces thus if its present?
Okay.So is it true to make an assumption that in banked curves the is never friction?,for some reason I cannot figure out since my statement is false.
 
  • #9
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"V=√rgtanθ is only valid when the car is not sliding I think,so this "v" will be the minimum speed.But for the upper limit I am not sure. "

The above formula is the speed when centrifugal force balances the force of the vehicle wanting to slide down the hill. It's the speed when there is no friction.

Determine the forces on the tires where they meet the pavement. You have 3 forces that you consider in order to determine the maximum speed given the banking and friction.

How do we know that a present centrifugal force in a particular problem balances the force of the vehicle wanting to slide down the hill.So that you do not include friction?


forces on the tires
Fnet(Vertical)=nsin(theta)-mg=0
Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
v=sqrt(grtan(theta)+rUs/sin(theta))
 
Last edited:
  • #10
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The formula V=sqrt(rg)tan(theta) is derived as follows.

Force up slope due to centrifugal force = mv^2/r * cos(theta)
Force down slope due to weight = mgsin(theta)

Equate them and you get the velocity for the zero friction case. You have friction in your problem so the force balance has another term.
 
  • #11
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I have added the extra term which is friction in my case.Is my equation correct now?
 
  • #12
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Where is the equation?
 
  • #13
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forces on the tires
Fnet(Vertical)=nsin(theta)-mg=0
Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
v=sqrt(grtan(theta)+rUs/sin(theta))
 
  • #14
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5
v=sqrt(grtan(theta)+rUs/sin(theta))

Are the units of gr the same as rUs? If not, they cannot be added and you have an error.
 
  • #15
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No.
Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
So I took the friction to be towards the center thus I added it to the horizontal forces.
 
  • #16
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Do you mind spotting my error because I cannot?
 
  • #17
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Just equate the forces up and down the slope. The component down the slope due to weight plus the friction force must equal to component of centrifugal force up the slope. That's the equation for maximum velocity.

Whenever deriving a formula, always check units. They will reveal errors if any exist. Correct units are a necessary but not sufficient condition for correctness.
 
  • #18
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OOOOOOhhhhh!....I see my mistake
Forces on the tires
Fnet(Vertical)=nsin(theta)-mg=0
Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
v=sqrt(grtan(theta)+grUs/sin(theta))
 
  • #19
1,198
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"v=sqrt(grtan(theta)+grUs/sin(theta))"

Look at your formula. Suppose the slope were zero. Your formula says the speed would be infinite. Your second term is the problem.
 
  • #20
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So along with unit checking, look at formulas you derive at the extremes to see if they provide realistic results and results that make sense.
 
  • #21
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Then I guess the maximum speed can be anything,you can turn safely at any speed.
But since you say its a "problem" then there is something I am missing here.
 
  • #22
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Yes, the maximum speed cannot be infinite. Try it with your Aston Martin and see what happens! You have an error in the last term. In the case of no slope, where theta is zero, the first term drops out as it should and you are left with only the friction term. Therefore the force balance must be such that the friction force equals the centrifugal force. Recheck your friction force.
 
  • #24
1,198
5
Looks correct

v=sqrt(grtan(theta)+grUs/sin(theta))

The above says that if the angle is zero, the speed is infinite. That is wrong because the vehicle can only go around the curve as fast as friction allows. With an angle of zero, the first term drops out which is correct but the second is infinite. And that is where the error lies.
 

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