How do I solve for the velocity in a banked curve with friction?

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Homework Help Overview

The discussion revolves around solving for the velocity of a vehicle navigating a banked curve with friction. Participants are exploring the dynamics involved, including forces acting on the vehicle and the role of friction in the context of circular motion.

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  • Mixed

Approaches and Questions Raised

  • Participants are attempting to derive equations for velocity while considering both the banking angle and friction. Some are questioning the assumptions made regarding forces, such as the normal force and its relationship to weight. Others are exploring how to express forces in terms of one another and the implications of acceleration in different directions.

Discussion Status

There is an ongoing exploration of the problem with various interpretations and approaches being discussed. Some participants have offered hints and guidance on how to approach the derivation of the equations, while others are reflecting on their understanding of the forces involved. A few have expressed moments of clarity, indicating progress in their understanding.

Contextual Notes

Participants are grappling with the complexities of the forces acting on a vehicle in motion on a banked curve, including the effects of friction and the need to account for different scenarios such as maximum and minimum speeds. There are indications of confusion regarding the normal force and its calculation in relation to the vehicle's motion.

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Vmax = 13.1 m/s
Vmin = 8.1 m/s

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I've spent all morning attempting part 4 of this problem :/ Can someone help me please?

FBD
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tanθ = v2 / rg
v = sqrt(rg)tanθ = sqrt [(20)(9.8)tan30] = 10.6376996 m/s
v = 10.6376996 m/s

Fdue to banked curve = mv2 / r = 5.658m

The centripetal force is the net force in the radial direction:
Fcentripetal = mv2 / r = mgμcosθ + 5.658m

Cancel the m's:
v2 /r = gμcosθ + 5.658
v2 / 20 = 9.8(0.2)cos30 + 5.658
v = 12.1 m/s??

Please help :/ thank you
 
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bulbasaur88 said:
tanθ = v2 / rg
v = sqrt(rg)tanθ = sqrt [(20)(9.8)tan30] = 10.6376996 m/s
v = 10.6376996 m/s
That equation is only good for the no friction case.

Hint: Apply Newton's 2nd law to derive a version that works for the friction case. (How is that equation derived? Same basic idea, only now you have an additional force.)

Hint 2: Your free body diagram shows only one of the conditions. How does it change for minimum speed versus maximum speed cases?
 
Hello Doc

I was trying to solve for the force of the banked curve separately from the force of friction... I eventually added them together...Did you see that I did that? Or is that just not the way to go about it?
 
Also, I understand that to find the other velocity, friction should be pointing the other way, right? I just did not bother to find this other value of v because I knew that my initial approach was incorrect...
 
bulbasaur88 said:
I was trying to solve for the force of the banked curve separately from the force of friction?? Did you see that I did that?
Yes, I saw what you tried to do but it didn't really make sense. What do you mean by the 'force of the banked curve'? The normal force?

In any case, to solve for the velocity you need to rederive the formula from scratch including friction.
 
bulbasaur88 said:
Also, I understand that to find the other velocity, friction should be pointing the other way, right?
Right!
 
Yes sir - I meant the normal force.

I will take a break, ponder on it, and take another stab at it tonight. I will get back to you. :)
 
Ack - I can't leave this problem alone...!


This is my second attempt. I feel that I am much closer, but still off somewhere.

Radial Direction
FNsin 30 + Fscos30 = mv2/r
mgcos30sin30 + mgμcos30 = mv2 / r

Drop the m's
gcos30sin30 + gμcos30 = v2/r

Solve for V
r[gcos30sin30 + gμcos30] = v2
v = sqrt(r[gcos30sin30 + gμcos30])
v = sqrt(20[9.8cos30sin30 + 9.8(0.2)cos30])
v = 10.9 m/s
 
bulbasaur88 said:
Radial Direction
FNsin 30 + Fscos30 = mv2/r
Good. But express Fs in terms of FN.
mgcos30sin30 + mgμcos30 = mv2 / r
Careful! You cannot assume that FN = mgcos30. That would be true if there were no component of acceleration normal to the surface, but that doesn't apply here.

Instead, leave FN as an unknown. Get a second equation for vertical forces.
 
  • #10
What is causing the acceleration normal to the surface? The bike only moves on the surface of the track. I do not see any movement in the direction of the normal force (AKA perpendicular to the surface of the track)...what I mean is, the bike isn't shooting upwards in the air in the direction of the normal force. So how can it have any type of acceleration in that way? I think not understanding this may be my problem. Is FN = mg?

Ok so for the horizontal forces I have:

FNsin30 + FNμcos30 = mv2 /r

With the velocity being in the x-direction.

For vertical forces I have:
FNcos30 - weight - Fssin30 = mv2 /r
FNcos30 - mg - FNμsin30 = mv2 /r

With the velocity in the y-direction.
 
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  • #11
bulbasaur88 said:
What is causing the acceleration normal to the surface? The bike only moves on the surface of the track. I do not see any movement in the direction of the normal force (AKA perpendicular to the surface of the track)...what I mean is, the bike isn't shooting upwards in the air in the direction of the normal force. So how can it have any type of acceleration in that way? I think not understanding this may be my problem.
The acceleration is horizontal. The normal to the surface is not vertical, so the acceleration has a component normal to the surface.

Is FN = mg?
No. (Why guess?)

Ok so for the horizontal forces I have:

FNsin30 + FNμcos30 = mv2 /r
Good.

With the velocity being in the x-direction.
Not sure what you mean here. The important thing is that the acceleration is horizontal.

For vertical forces I have:
FNcos30 - weight - Fssin30 = mv2 /r
FNcos30 - mg - FNμsin30 = mv2 /r

With the velocity in the y-direction.
Why are you setting vertical forces equal to mv2/r? There's no vertical acceleration.
 
  • #12
I just noticed that I totally neglected the force of mgsin30 down the slope this entire time.
 
  • #13
Well, that didn't help. Haha. I admit defeat on this problem. It's like when you say a word too many times...this problem has lost all meaning.
 
  • #14
HOLY MOLY...I JUST GOT IT. I had to piece some of the vague hints you gave in earlier posts together and WHAM. THANK YOU DOC AL! you never fail me. haha I've never worked on a single problem this long :))) feels so rewarding :)))
 
  • #15
Although I've reached the correct answer, I still do not see why FN does NOT equal mgcos30.
 
  • #16
bulbasaur88 said:
Although I've reached the correct answer, I still do not see why FN does NOT equal mgcos30.

If you drive a car, at speed, through a dip in the road, the car's suspension is compressed and the passengers feel they have been pushed down into the seats. That is because at the bottom of the dip, the FN has to not only balance weight, but provide the necessary [upward] centripetal force so is much stronger than is was when driving on flat ground.

When you drive a car around a banked curve, the FN is also larger than that needed if the car was stationary - for the same reason. [of couse if the car was stationary you would need some friction to prevent the car slipping down the banking]
 
  • #17
awesome thank you peterO :)
 

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