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

In summary: I'm sorry, I don't understand what you're trying to say. The force of mgsin30 causes the bike to accelerate down the slope.
  • #1
bulbasaur88
57
0
Untitled.png


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

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

FBD
Untitled2.png


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
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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?
 
  • #4
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...
 
  • #5
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.
 
  • #6
bulbasaur88 said:
Also, I understand that to find the other velocity, friction should be pointing the other way, right?
Right!
 
  • #7
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. :)
 
  • #8
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
 
  • #9
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.
 
Last edited:
  • #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 :)
 

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

What is a banked curve?

A banked curve is a type of curve on a surface where the surface slopes upwards on the outer edge of the curve and downwards towards the center. This allows for a more natural and safer path for an object to follow, such as a car or rollercoaster.

How does banking a curve affect friction?

Banking a curve can reduce the amount of friction between the object and the surface, allowing for a smoother and faster movement. The slope of the curve counteracts the centrifugal force and provides a normal force that helps the object maintain its speed and direction.

What factors affect the amount of friction on a banked curve?

The angle of the banking, the speed of the object, and the coefficient of friction between the object and the surface are all factors that can affect the amount of friction on a banked curve. A higher angle of banking and a higher coefficient of friction will result in more friction, while a lower speed will result in less friction.

Can a banked curve be too steep?

Yes, a banked curve can be too steep. If the angle of banking is too high, it can cause the object to slide or lose control. This is why engineers carefully calculate and design the angle of banking for curves on racetracks and rollercoasters to ensure the safety and stability of the object.

How does friction affect the speed of an object on a banked curve?

Friction can affect the speed of an object on a banked curve in different ways depending on the angle of banking. If the angle is too low, friction can cause the object to slow down. However, if the angle is just right, friction can help the object maintain its speed and prevent it from sliding off the curve.

Similar threads

Solving Banked Curve Problem: Coefficient of Static Friction
Replies
6
Views
3K
Finding max velocity for a kart on a circular, banked track
Replies
2
Views
1K
Find the Coefficient of friction on a banked curve.
Replies
9
Views
6K
Banking a Curve without Friction
Replies
16
Views
9K
Maximum Speed on a Banked Curve with Friction
Replies
19
Views
3K
Banked curve with friction, but no angle
Replies
2
Views
3K
Find Banking Angle and Normal/Friction Force on Curved Road
Replies
4
Views
9K
Solving Banked Curve Friction Problems
Replies
6
Views
3K
Finding friction coefficient, car unbanked curve.
Replies
3
Views
3K
What Minimum Coefficient of Static Friction Prevents Skidding on a Banked Curve?
Replies
1
Views
3K

Hot Threads

Recent Insights

Back
Top