Maximum distance the car accelerates in a circle without skidding

In summary: 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?/Car's engine isn't 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
  • #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?
 
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  • #2
I think you have missed some syntax in your LaTex formatting.
 
  • #3
Did you mean so?
Pushoam said:

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.
 
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  • #4
ehild said:
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?
 
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  • #5
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.
 
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  • #6
  • #7
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
ehild said:
Did you mean so?
Yes, thanks for it.

scottdave said:
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?
Hiero said:
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.

haruspex said:
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
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
scottdave said:
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
Pushoam said:
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.
 
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  • #12
haruspex said:
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
Pushoam said:
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
haruspex said:
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
Pushoam said:
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
haruspex said:
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
Pushoam said:
If all of the accelerations come from the tyre/road contact, gravity and air resistance, then why do we need an engine?
Think about what the engine does. It exerts a torque on the axle. Without friction on the road, that would cause the wheel to spin. The friction opposes the tyre sliding on the road, and that pushes the car along.
See if https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/ helps.
 
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1. What is the maximum speed at which a car can accelerate in a circle without skidding?

The maximum speed at which a car can accelerate in a circle without skidding depends on several factors, including the car's weight, tire traction, and the radius of the circle. However, in general, the maximum speed is determined by the equation v = √(μrg), where μ is the coefficient of friction, r is the radius of the circle, and g is the acceleration due to gravity.

2. How does the weight of the car affect its maximum acceleration in a circular motion?

The weight of the car has a significant impact on its maximum acceleration in a circular motion. A heavier car will have a higher maximum speed at which it can accelerate without skidding compared to a lighter car. This is because a heavier car has more force pressing down on the tires, increasing the friction between the tires and the road. This increased friction allows the car to maintain a higher speed without skidding.

3. What is the significance of tire traction in determining the maximum acceleration in a circular motion?

Tire traction plays a crucial role in determining the maximum acceleration in a circular motion. It refers to the amount of grip that the tires have on the road. The higher the tire traction, the more force can be applied to the tires without them slipping or skidding. Therefore, a car with tires that have a higher traction will be able to accelerate at a higher speed in a circular motion without skidding.

4. How does the radius of the circle affect the maximum acceleration of a car without skidding?

The radius of the circle also plays a significant role in determining the maximum acceleration of a car without skidding. The larger the radius, the lower the acceleration needed to maintain a certain speed. This means that a car can accelerate at a higher speed without skidding on a larger circle compared to a smaller one. Additionally, a larger radius also allows for a smoother turn, reducing the chances of skidding.

5. What is the coefficient of friction and how does it affect the maximum acceleration in a circular motion?

The coefficient of friction is a measure of the amount of friction between two surfaces. In the context of a car accelerating in a circle, it refers to the friction between the tires and the road. A higher coefficient of friction means that there is more grip between the tires and the road, allowing for a higher maximum acceleration without skidding. It is affected by factors such as tire material, road surface, and weather conditions.

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