Centripetal Force on a banked road

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Homework Help Overview

The discussion revolves around determining the angle at which a road should be banked for a racing car to achieve maximum control, given specific parameters such as radius, speed, and coefficient of friction. The subject area includes concepts of centripetal force and friction in the context of circular motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between banking angle, friction, and centripetal force. There are attempts to derive equations involving normal force and friction, with some questioning the inclusion of friction in the analysis for maximum control.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some suggest that the problem may be poorly constructed, while others emphasize the need for clarity on the role of friction and the conditions for maximum control.

Contextual Notes

There is uncertainty regarding the complete problem statement and whether it is part of a multi-part question. Participants express confusion over the implications of the coefficient of friction provided in the problem.

spartacus
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Homework Statement


50.5 m radius
19 m/s speed
µ = 0.2
acceleration due to gravity is 9.8 m/s2
At what angle θ should it be banked for
maximum control of the racing car?

Homework Equations


sum of total forces=0?

The Attempt at a Solution


N*cosθ=mg
N*sinθ-mgµ=(mv^2)/r
mg*tanθ-mgµ=(mv^2)/r
tanθ-µ=v^2/r
θ=arc tan((v^2/r)+µ)
θ=82.25 degrees?
I feel like i did everything right, seems high though...
 
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spartacus said:
sum of total forces=0?
No, it's accelerating.

If you want to solve for 'maximum control', I would ignore friction. But if you do include friction, which way will it act?
 
Doc Al said:
No, it's accelerating.

If you want to solve for 'maximum control', I would ignore friction. But if you do include friction, which way will it act?
Sorry, i meant that as a relevant equation in one dimension, since it's not accelerating vertically
 
spartacus said:
Sorry, i meant that as a relevant equation in one dimension, since it's not accelerating vertically
Sure, the sum of the vertical force components must be zero.
 
Does maximum control mean with aid of friction?

Are N*sinθ and mgµ acting along the same line?
 
Doc Al said:
Sure, the sum of the vertical force components must be zero.
ya so could you help me see what i did wrong?
 
Jazz said:
Does maximum control mean with aid of friction?

Are N*sinθ and mgµ acting along the same line?
I figured N*sinθ is equal to centripetal force so it would by definition be going in the opposite direction of friction
 
spartacus said:
ya so could you help me see what i did wrong?
For one thing, you included friction in your analysis. For another, you failed to consider the direction that friction acts.
 
spartacus said:
I figured N*sinθ is equal to centripetal force
In the absence of friction.

spartacus said:
so it would by definition be going in the opposite direction of friction
What's the direction of the centripetal acceleration? What's the direction of friction with respect to the surface on which it acts?
 
  • #10
Doc Al said:
For one thing, you included friction in your analysis. For another, you failed to consider the direction that friction acts.
Why wouldn't i include friction?
Didnt i consider the direction being against the horizontal normal force by subtracting it in N*sinθ-mgµ=(mv^2)/r
 
  • #11
spartacus said:
Why wouldn't i include friction?
Because it asks for the angle of 'maximum control'. Is this question a part of a multi-part question?

spartacus said:
Didnt i consider the direction being against the horizontal normal force by subtracting it in N*sinθ-mgµ=(mv^2)/r
Are you assuming that friction acts horizontally?
 
  • #12
Doc Al said:
Because it asks for the angle of 'maximum control'. Is this question a part of a multi-part question?Are you assuming that friction acts horizontally?
I believe the question means maximum control on the road that has the coefficient of friction of .2 or else why would mu be provided
and yes
 
  • #13
Please don't give up on me, i easily did the other questions from this problem set, this question shouldn't be too hard, i just don't know what i did wrong
 
  • #14
spartacus said:
I believe the question means maximum control on the road that has the coefficient of friction of .2
Note that when you consider friction, you have a range of acceptable speeds. Friction will give you the maximum and minimum values of the speeds that can be maintained without slipping. I would interpret the angle of maximum control to be that angle where friction is not needed.

The problem could just be poorly constructed. Are you posting the complete problem, word for word?

spartacus said:
or else why would mu be provided
Because this could be the first part of a multipart problem. Is it?

spartacus said:
and yes
That is incorrect. Friction acts parallel to the surface.
 
  • #15
spartacus said:
Please don't give up on me
I'm not giving up on you. :)

The first thing is to make sure we have the complete problem stated. Is this from a textbook? Or devised by the instructor?
 
  • #16
Doc Al said:
Note that when you consider friction, you have a range of acceptable speeds. Friction will give you the maximum and minimum values of the speeds that can be maintained without slipping. I would interpret the angle of maximum control to be that angle where friction is not needed.

The problem could just be poorly constructed. Are you posting the complete problem, word for word? Because this could be the first part of a multipart problem. Is it?That is incorrect. Friction acts parallel to the surface.
It's not a multipart problem no, how would i incorporate force of friction? would it be mg sintheta-mu?
 
  • #17
Doc Al said:
I'm not giving up on you. :)

The first thing is to make sure we have the complete problem stated. Is this from a textbook? Or devised by the instructor?
i have it right here let ill upload a screenshot
 
  • #18
1212.png
 
  • #19
It's a bit perplexing as to what they mean by 'maximum control'. I would interpret it as the angle that requires no friction.

Is this from some textbook?
 
  • #20
Doc Al said:
It's a bit perplexing as to what they mean by 'maximum control'. I would interpret it as the angle that requires no friction.

Is this from some textbook?
No, it's an online assignment

The way i solved it, i just used the data to find which angle it would be at that speed
 
  • #21
If you're going to apply friction, realize that the maximum value is given by ##\mu N## and that it acts parallel to the surface, not horizontally. And it could act up the incline or down the incline, depending on the angle.
 
  • #22
Doc Al said:
If you're going to apply friction, realize that the maximum value is given by ##\mu N## and that it acts parallel to the surface, not horizontally. And it could act up the incline or down the incline, depending on the angle.
If it's parallel should i use mu*mg sintheta then because that would make the problem a lot more difficult
 
  • #23
spartacus said:
No, it's an online assignment
It seems like a poorly worded problem to me.
 
  • #24
spartacus said:
If it's parallel should i use mu*mg sintheta then
The normal force does not equal mg.
 
  • #25
I would solve the problem assuming no friction. Perhaps they are just messing with your head to see if you know what matters and what doesn't. ;)
 
  • #26
Doc Al said:
The normal force does not equal mg.
i mean having Ncostheta=mg for the vertical and Nsintheta-mu*mg*sintheta=(mv^2)/r for horizontal
 
  • #27
Doc Al said:
I would solve the problem assuming no friction. Perhaps they are just messing with your head to see if you know what matters and what doesn't. ;)
lol, i'll definitely get clarification from my teacher monday morning
 
  • #28
spartacus said:
i mean having Ncostheta=mg for the vertical and Nsintheta-mu*mg*sintheta=(mv^2)/r for horizontal
Two problems. Since friction has both vertical and horizontal components, your first equation would have to be modified. And, once again, the normal force does not equal mg.
 
  • #29
spartacus said:
lol, i'll definitely get clarification from my teacher monday morning
Good idea!
 

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