Circular Motion, Banked Turn Without Friction

In summary: If you don't want the maximum radius try this:Draw a 36×36 square, the left side & the top along the two streets. The diagonal from the upper right to the lower left, just touches Bubba's. Draw the upper left quarter of the circle with radius 36 meters and centered at the lower right corner of this square. It easily misses Bubba's.
  • #1
mobwars
9
0
1.Traffic travels north on Newton blvd and turns east onto the road einstein way. The speed of the traffic traveling north and east is 55 mph (24.6m/s). The banked turn must fit within an area (shown in the picture attached). The illustration is not drawn to scale.

You are to design an exit ramp which is banked such that it requires no friction in the radial direction. Also, you are to design the ramp so the vehicles maintain their speed and are not required to slow down.

2. Fnet=ma, r=v^2/(g*tan(theta)), r=(v*T)/(2*pi), v^2=g*r*tan(theta)


3. What I have attempted so far is this:

First I set r equal to v^2/(g*tan(theta)), and r equal to (v*T)/(2*pi), and then set the two equations equal to each other. My thought process is that since there are two unknowns, the radius and the angle of the banked turn, I should set up a system of equations in order to solve for one variable so that I can find the other. The point in time where I run into trouble is when I don't know the period of time to make one revolution (T).

Am I making a dumb mistake, or am I using the wrong equations? Also, is it safe to assume that, since the vehicles will stay at a constant velocity of 24.6 m/s, that the velocity given is also the critical velocity?

http://s296.photobucket.com/albums/mm179/mobwars911/?action=view&current=PHYSICS.jpg"
 
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  • #2
Get the maximum r from analyzing the figure. Maybe give Bubba a little extra room.

You don't need to find T.

Assuming your equations are for the situation shown here, find θ once you get r.
 
  • #3
SammyS said:
Get the maximum r from analyzing the figure. Maybe give Bubba a little extra room.

You don't need to find T.

Assuming your equations are for the situation shown here, find θ once you get r.

Okay. I analyzed the figure and I'm still a little confused. The maximum r would be found by seeing that the dimensions make an imaginary square 18m x 18m. I'm just so lost.

I know this isn't for just getting the answer, but I've been staring at this picture for past hour. I know I'm over thinking it.
 
  • #4
Here's what I have thought of as far as maximum radius is concerned.

If the space above bubba's is 18 m and the space beside bubba's is 18m, then the two dimensions make a triangle (in the form of a^2+b^2=c^2) (18+x)^2+(18+x)^2=c^2. Am I on the right track?
 
  • #5
Finding r is basically a geometry or a trigonometry problem

Inscribe a circle in a square. The left side & top side of the square represent the two streets. The upper left 1/4 of the circle represents the ramp.

Now inscribe a smaller square in the circle. Bubba's BBQ sits in the upper left corner of this smaller square. The two squares are separated by 18 meters.

Find the radius of the circle.
 
  • #6
If you don't want the maximum radius try this:

Draw a 36×36 square, the left side & the top along the two streets. The diagonal from the upper right to the lower left, just touches Bubba's. Draw the upper left quarter of the circle with radius 36 meters and centered at the lower right corner of this square. It easily misses Bubba's.
 
  • #7
If you don't want the maximum radius try this:

Draw a 36×36 square, the left side & the top along the two streets. The diagonal from the upper right to the lower left, just touches Bubba's. Draw the upper left quarter of the circle with radius 36 meters and centered at the lower right corner of this square. It easily misses Bubba's.
 
  • #8
SammyS said:
If you don't want the maximum radius try this:

Draw a 36×36 square, the left side & the top along the two streets. The diagonal from the upper right to the lower left, just touches Bubba's. Draw the upper left quarter of the circle with radius 36 meters and centered at the lower right corner of this square. It easily misses Bubba's.

Yes, okay thank you. That makes so much more sense. This seems like a problem that you have to be able to visualize more than anything else, and the way you worded that make it extremely clear. Thank you thank you thank you. Have a great night!
 
  • #9
You're welcome & good luck.
 
  • #10
SammyS said:
You're welcome & good luck.

Everything else is going smoothly, as I outlined every other process that I would need to go through for the later calculations. Pretty insane angle for the banked turn though...59.8 degrees. Certainly would be interesting to try and drive on that.
 
  • #11
Even at the very maximum radius, I get about 45°.

The Daytona Speedway's turns are only banked at 31°, and that is very extreme.
 
  • #12
SammyS said:
Even at the very maximum radius, I get about 45°.

The Daytona Speedway's turns are only banked at 31°, and that is very extreme.

Hmm...what equation did you use?

I used (theta)=tan^-1(v^2/rg)
 
  • #13
Your answer is fine.

It's possible to use a radius of about 61 meters. That gives about 45°.

I don't know what figures Bill France used at Daytona. LOL
 
  • #14
SammyS said:
Your answer is fine.

It's possible to use a radius of about 61 meters. That gives about 45°.

I don't know what figures Bill France used at Daytona. LOL

radius MUST be less than 62 right?

EDIT: and indeed. lol.
 
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  • #15
Yes, a little less than 62. Actually the max radius is:

[tex]r_{max}=\left(2+\sqrt{2}\,\right)\cdot18[/tex]
 
  • #16
SammyS said:
Yes, a little less than 62. Actually the max radius is:

[tex]r_{max}=\left(2+\sqrt{2}\,\right)\cdot18[/tex]

okay, cool. thank you again for all of this, you've been a great help!
 

1) What is circular motion and how is it related to a banked turn without friction?

Circular motion refers to the motion of an object along a circular path. In a banked turn without friction, the object is moving along a curved path without any frictional force acting on it. This results in the object following a circular path.

2) How does the banked angle affect circular motion in a banked turn without friction?

The banked angle, also known as the angle of incline, is the angle at which the curved surface is tilted. The banked angle affects the circular motion by providing a centripetal force that keeps the object moving in a circular path without the need for friction. The greater the banked angle, the greater the centripetal force.

3) What is the role of centripetal force in a banked turn without friction?

Centripetal force is the force that keeps an object moving in a circular path. In a banked turn without friction, the centripetal force is provided by the horizontal component of the object's weight, which is directed towards the center of the circle. This force acts as the centripetal force and keeps the object in circular motion.

4) Are there any other factors that affect circular motion in a banked turn without friction?

Apart from the banked angle, the speed of the object also affects circular motion in a banked turn without friction. As the speed increases, the centripetal force required to keep the object in circular motion also increases. This means that a higher banked angle is needed for a higher speed in order to maintain a banked turn without friction.

5) Can a banked turn without friction be achieved in real-world situations?

In theory, a banked turn without friction can be achieved, but it is difficult to achieve in real-world situations. In order to have a true banked turn without friction, the curved surface must be perfectly smooth and the object must have enough speed to create the necessary centripetal force. In real-world situations, there is almost always some amount of friction present, which makes it difficult to achieve a banked turn without friction.

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