[Guest]

I was remembering back into when I first learned physics I (kinematics) and I remembered a problem that at first gave me some trouble.

A car on a race track has a turn that is a perfectly circular pi radian turn with radius 20 m. The track is angled at theta. (this angle is the angle the track makes with a perfectly horizontal base in the direction of the perfectly vertical). You have to figure out the maximum speed that is possible on the turn. For this problem you want to find the equilibrium point so that the car does not move vertically up or down the track.

Three questions (for gravity use 9.81)

1. Without friction on the track what is the ideal speed if theta is pi/18?

2. What about if you have a coefficient of static friction of .3 with angle pi/18?

now when i went back to these problems i realized they weren't as hard as they seemed in highschool. so I thought of another problem and have not solved it yet (cant think of how to proceed from where im at, but i haven't tried a linear example) , what if the angle was variable. For example a louge flat on the straight away but banks on the turn.

Now to simplify so the equation of the angle can be linear and not make things as complicated, the turn is now only pi/2 radians. So say the function that defines theta is theta=g(x)=(x/90) with x equaling the distance traveled in meters. What is the time it takes to make this quarter circle turn. for simplicity imagine that the car can be at any speed instantaneously and acceleration in the direction of your velocity is zero. the coefficient of static friction is still .3

[bonanova]

[1] 5.73 m/s and [2] 9.67 m/s

v² = gr tan(a) where

[1] a = pi/18 and [2] a= pi/18 + tan^-1(.3)

Clarification?

" For example a louge flat on the straight away ... " ?

[bonanova]

... what if the angle was variable.

For example a louge flat on the straight away but banks on the turn.

Now to simplify so the equation of the angle can be linear and not make things as complicated, the turn is now only pi/2 radians.

So say the function that defines theta is theta=g(x)=(x/90) with x equaling the distance traveled in meters.

What is the time it takes to make this quarter circle turn.

for simplicity imagine that the car can be at any speed instantaneously and acceleration in the direction of your velocity is zero.

the coefficient of static friction is still .3

Is this the third question?

1. Theta = x/90 radians, where x is distance in meters from the start of the turn.

2. Turn radius is 20m.

3. Velocity is constant [acceleration is 0.]

4. We're being asked how high that velocity can be, given .3 for friction.

If so, it's just question 2 for theta = pi/9 [the final angle: x = 20 pi/2]

Also, requiring acceleration to be zero doesn't gibe with allowing the velocity to change instantaneously.

I think I misinterpreted something.

[Guest]

basically you got it right but it is a common formula to use a=v²/r where A is acceleration a vector orthogonal to the velocity. This if there is constant speed(not velocity). I am saying the angle of the road changes and this ideal speed changes to. If you can magically be going this ideal speed at every moment what is the time it takes. I simply meant you did not need to use acceleration in the direction of your velocity (though you can) you could consider the speed magically optimal at all times.

so for number 3(which i accidentally never numbered)

r=20m

theta=x/90

x is distance traveled from start of problem

so at the end of this problem which really is the middle of a turn your at

theta=(2pir/4)/90=10pi/90=pi/9

mu =.3

as for louge i meant ice luge it was the best example I had

[HoustonHokie]

basically you got it right but it is a common formula to use a=v²/r where A is acceleration a vector orthogonal to the velocity. This if there is constant speed(not velocity). I am saying the angle of the road changes and this ideal speed changes to. If you can magically be going this ideal speed at every moment what is the time it takes. I simply meant you did not need to use acceleration in the direction of your velocity (though you can) you could consider the speed magically optimal at all times.

so for number 3(which i accidentally never numbered)

r=20m

theta=x/90

x is distance traveled from start of problem

so at the end of this problem which really is the middle of a turn your at

theta=(2pir/4)/90=10pi/90=pi/9

mu =.3

as for louge i meant ice luge it was the best example I had

Actually, this problem is not that bad. The easiest way I found to do it was to break the track into very short pieces, figure out the velocity at the beginning and end of each piece, take the average and figure out the drive time at the average speed for each section. You could also set up an integral function, but I'm too lazy and I like spreadsheets too much. Minimum speed at the beginning is 7.67 m/s and maximum speed at the end is 12.09 m/s. All totaled, I got 3.22 s to travel the 31.4 m quarter turn.

As an aside (and maybe even a rabbit trail), I assume that you're not actually coming off of a straight piece of track, but rather a curved piece with no banking. If you were, the theoretical maximum speed on the straight piece is infinite and the deceleration at the start of the curve would kill the driver. Assuming the infinite speed didn't get to him first...

The other problem would be a physical inability to turn the wheels instantaneously. You actually drive in a spiral when going from straight to curved sections of road to avoid a sudden change in centripital force (a=v²/r). If r goes from infinity on a straight line to, say, 20 m in the curve, you have an instantaneous change in acceleration, which is not good for your vehicle . That's why railroads layout their curves with clothoid spirals at the beginning and end, or else the trains would jump the tracks. Some highways are laid out that way as well, but most have plain old curves because drivers make the spiral themselves.

[Guest]

i realized alot of the inconsistencies with this problem but i found the integral and was having trouble solving it (i normally do when theres trigonometry and quotient). That was when the equation for the angle wasn't linear though i in all honesty havent done these problems with these values. Anyway I realized that the value at the beginning is infinite and was going to impose a max velocity on the car but then forgot after it took longer to explain the concept then normal.

and as for the last part i dont know if you mean the bank in the roads or the turn, but i dont beleive this is harder to accept then the lack of forward acceleration while magically being at different velocities. I simply remembered that old problem and then thought of a what if problem and here we are. Anyway Ill solve these later.