Maximum distance the car accelerates in a circle without skidding

Click For Summary
SUMMARY

The maximum distance a car can accelerate in a circle without skidding is determined by the relationship between static friction, centripetal acceleration, and tangential acceleration. The derived formula for maximum distance, denoted as s_{max} = \frac{kgR}{2w_τ}, incorporates the static friction coefficient, radius, and angular acceleration. Key insights include the necessity of considering both centripetal and tangential forces, as static friction provides the required accelerations. The discussion emphasizes the importance of using static friction rather than kinetic friction in calculations related to circular motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with concepts of centripetal and tangential acceleration
  • Knowledge of static versus kinetic friction
  • Basic proficiency in algebra and physics equations
NEXT STEPS
  • Study the principles of circular motion in physics
  • Learn about the role of friction in dynamics, focusing on static friction
  • Explore the derivation of equations related to centripetal acceleration
  • Investigate real-world applications of friction in automotive engineering
USEFUL FOR

Students studying physics, automotive engineers, and anyone interested in understanding the dynamics of vehicles in circular motion.

Pushoam
Messages
961
Reaction score
53

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?
 
Physics news on Phys.org
I think you have missed some syntax in your LaTex formatting.
 
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.
 
Last edited:
  • Like
Likes   Reactions: scottdave
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?
 
  • Like
Likes   Reactions: haruspex
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.
 
  • Like
Likes   Reactions: scottdave
There is a mistake in the question. It should provide the static friction coefficient, not that of "sliding" (kinetic) friction.
 
  • Like
Likes   Reactions: scottdave and Chestermiller
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?
 
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.
 
  • Like
Likes   Reactions: Pushoam
  • #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.
 
  • Like
Likes   Reactions: Pushoam

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K