Centripetal Force on a banked road

[spartacus]

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...

 

[Doc Al]

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?

 

[spartacus]

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

 

[Doc Al]

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.

 

[Jazz]

Does maximum control mean with aid of friction?

Are N*sinθ and mgµ acting along the same line?

 

[spartacus]

Sure, the sum of the vertical force components must be zero.

ya so could you help me see what i did wrong?

 

[spartacus]

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

 

[Doc Al]

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.

 

[Doc Al]

I figured N*sinθ is equal to centripetal force

In the absence of friction.

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?

 

[spartacus]

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

 

[Doc Al]

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?

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?

 

[spartacus]

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

 

[spartacus]

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

 

[Doc Al]

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?

or else why would mu be provided

Because this could be the first part of a multipart problem. Is it?

and yes

That is incorrect. Friction acts parallel to the surface.

 

[Doc Al]

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?

 

[spartacus]

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?

 

[spartacus]

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

 

[spartacus]

 

[Doc Al]

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?

 

[spartacus]

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

 

[Doc Al]

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.

 

[spartacus]

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

 

[Doc Al]

No, it's an online assignment

It seems like a poorly worded problem to me.

 

[Doc Al]

If it's parallel should i use mu*mg sintheta then

The normal force does not equal mg.

 

[Doc Al]

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. ;)

 

[spartacus]

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

 

[spartacus]

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

 

[Doc Al]

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.

 

[Doc Al]

lol, i'll definitely get clarification from my teacher monday morning

Good idea!