Maximum distance the car accelerates in a circle without skidding

  • #1
Pushoam
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Homework Statement


upload_2017-7-20_20-27-10.png


Homework Equations




The Attempt at a Solution


At distance s, the speed of the car is v.
$$ v^2 = 2wτs$$
$$\frac { mv^2} R ≤ kmg$$
Let's denote the maximum distance covered without sliding is smax.
$$\frac { m2wτsmax} R = kmg$$
$$ smax = \frac {kgR} {2wτ}$$

Is this correct so far?
 

Answers and Replies

  • #2
cnh1995
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I think you have missed some syntax in your LaTex formatting.
 
  • #3
ehild
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Did you mean so?

Homework Statement


View attachment 207523

Homework Equations




The Attempt at a Solution


At distance s, the speed of the car is v.
$$ v^2 = 2w_τ s$$
$$\frac { mv^2} {R} ≤ kmg$$
Let's denote the maximum distance covered without sliding is ##s_{max}##.
$$\frac { m2w_τ s_{max} }{R} = kmg$$
$$ s_{max} = \frac {kgR} {2w_τ}$$

Is this correct so far?
It is correct.
 
Last edited:
  • #4
scottdave
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Did you mean so?
It is correct.
But should you also consider the tangential acceleration and take the vector sum of that and the centripetal acceleration, when calculating the horizontal force on the car?
 
  • #5
Hiero
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I agree with @scottdave; if we assume that the tangential acceleration is due to the cars engine (and not, say, someone pushing the car) then the tangential acceleration must also be supplied by the frictional force, and so @Pushoam, your inequality is not quite right.
 
  • #7
haruspex
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There is a mistake in the question. It should provide the static friction coefficient, not that of "sliding" (kinetic) friction.
 
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  • #8
Pushoam
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Did you mean so?
Yes, thanks for it.

But should you also consider the tangential acceleration and take the vector sum of that and the centripetal acceleration, when calculating the horizontal force on the car?
if we assume that the tangential acceleration is due to the cars engine (and not, say, someone pushing the car) then the tangential acceleration must also be supplied by the frictional force, and so @Pushoam, your inequality is not quite right.

Earlier, I had seen that in a uniform circular motion, when there is no force except friction acts on the body, the friction force provides the centripetal acceleration.

Here, you say that there is a component of friction in centripetal direction providing centripetal acceleration and another component in tangential direction.
But, why should one component of friction act in tangential direction?
It is not said in the question that the friction force provides tangential acceleration.

There is a mistake in the question. It should provide the static friction coefficient, not that of "sliding" (kinetic) friction.
We are going to write the eqn. of motion when the body is moving, so,the question should provide kinetic friction. Isn't it so?
 
  • #9
scottdave
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Car tires work by maintaining static friction with the pavement for the small patch of tire in contact with the road. If there is no friction, or if you are sliding then the tires are skidding. So yes the friction is causing the car to acelerate.
 
  • #10
Pushoam
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So yes the friction is causing the car to acelerate.
The friction is causing the car to have centripetal acceleration. This I understood.
Do you mean that the friction is causing the car to have tangential acceleration, too?
And this tangential acceleration is wτ.
 
  • #11
haruspex
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The friction is causing the car to have centripetal acceleration. This I understood.
Do you mean that the friction is causing the car to have tangential acceleration, too?
And this tangential acceleration is wτ.
The friction (static here) is providing the net acceleration, so both the centripetal and the tangential.
Static friction acts to oppose commencement of relative motion of the surfaces in contact; kinetic friction opposes actual relative motion.
Consequently, if the friction were kinetic it would be antiparallel to the relative velocity, so subsequent motion would be in a straight line. Kinetic friction will not get you round a bend.
 
  • #12
Pushoam
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The friction (static here) is providing the net acceleration, so both the centripetal and the tangential.
How do we get to know this ?
How will I find out in which case friction provides only centripetal acceleration and in which case it provides both centripetal and tangential accelerations?
 
  • #13
haruspex
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How do we get to know this ?
How will I find out in which case friction provides only centripetal acceleration and in which case it provides both centripetal and tangential accelerations?
Friction is the only horizontal force acting. ##\Sigma\vec F=m\vec a##, where ##\vec a## is the acceleration of the mass, not just some component of it.
 
  • #14
Pushoam
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Friction is the only horizontal force acting.
Car's engine couldn't provide centripetal acceleration, but, it could provide tangential acceleration. I am asking this because earlier I assumed that car's engine provides tangential acc. and friction provides centripetal acc.

So, if the question doesn't specify that it's car's engine which provides tan. acc. , then, we have to take friction providing tan. acc. Is it so?/
 
  • #15
haruspex
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Car's engine couldn't provide centripetal acceleration, but, it could provide tangential acceleration.
The car's engine is not in contact with the road. The car only has acceleration in consequence of external forces acting on it, and those all come via the tyre/road contact, gravity and air resistance. Without friction between tyre and road, the engine won't get you anywhere.
 
  • #16
Pushoam
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The car only has acceleration in consequence of external forces acting on it, and those all come via the tyre/road contact, gravity and air resistance.
If all of the accelerations come from the tyre/road contact, gravity and air resistance, then why do we need an engine?
 
  • #17
haruspex
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