Simple banked curve, centripetal acceleration problem

[bulbasaur88]

If a car can negotiate a curve of radius 200 m at 30 m=s without any frictional force being
required, what is the banking angle

r = 200 m
v = 30 m/s

tanΘ = v²/r
Θ = tan^-1(v²/r]
Θ = tan^-1(30²/200]
Θ =77.47 degrees

I thought this problem was a simple one, but the answer that is provided is Θ = 27 degrees. Will somebody tell me if I did it wrong? Like I said, I thought it was a pretty straightforward question...Hrmm.. Thank you in advance :)

 

[Pi-Bond]

I think you got the centripetal force wrong. Draw a free body diagram and see which force causes the circular motion.

 

[PeterO]

If a car can negotiate a curve of radius 200 m at 30 m=s without any frictional force being
required, what is the banking angle

r = 200 m
v = 30 m/s

tanΘ = v²/r
Θ = tan^-1(v²/r]
Θ = tan^-1(30²/200]
Θ =77.47 degrees

I thought this problem was a simple one, but the answer that is provided is Θ = 27 degrees. Will somebody tell me if I did it wrong? Like I said, I thought it was a pretty straightforward question...Hrmm.. Thank you in advance :)

Notice that you have not used gravity at any stage, so you are claiming we need exactly the same banking on the Earth, Moon, Jupiter. your v^2 / R value is fine.

 

[Lobezno]

See my post below for the correct response.

 

[Pi-Bond]

You don't need the mass, it will cancel when equating forces.

 

[PeterO]

I'm not aware (not that it doesn't exist, simply I'm not aware of it) of any way one would solve it that didn't require the mass of the car.

When a "roller coaster" goes around banked curve, you don't have the light people fall in one direction and the heavy people fall in the other! Everyone - including the empty cars - goes around the curves just fine.

The banking required is independent of mass -that is why they can design amusement park rides - but it is not independent of the acceleration due to gravity in the region of the amusement park. Fortunately g = 9.8 covers all our amusement parks.

It also allows road builders to decide what degree of banking they should use for a curve on the highway, without knowing if you will be driving a 900kg small car or a 3 tonne Rolls Royce.
On the highway, they have to base their calculations on a "typical" speed of cars on that section of road, and friction between the tyres and the road becomes far more critical. That also explains why more people run off the road when the road is wet or covered in mud [the co-efficient of friction is smaller]

 

[PeterO]

You don't need the mass, it will cancel when equation forces.

Notice that weight is mg and centripetal force is mv^2 / R.

Both forces are proportional to mass, so the ratio of the two is independent of mass.

 

[Lobezno]

Oooooooooops yeah I realized that a few minutes later. My bad!
cos(theta)=(v^2)rg not tan. Like said above, draw the free body diagram and you'll see why.

cos^1(4.5/g)=theta -> use as many decimal places of g as you please. I can never be bothered to do it with anything other than g=10, but I generally do these problems without paper.

 

[PeterO]

Oooooooooops yeah I realized that a few minutes later. My bad!
cos(theta)=(v^2)r not tan. Like said above, draw the free body diagram and you'll see why.

cos^1(4.5/g)=theta -> use as many decimal places of g as you please. I can never be bothered to do it with anything other than g=10, but I generally do these problems without paper.

That cos(theta)=(v^2)r should be cos(theta)=(v^2)rg, but you put the g in on the next line.

 

[Lobezno]

A typo, but yeah, it's in the next line.

 

[bulbasaur88]

Whoops I forgot gravity. However with gravity included my answer is still not right :/

MY RE-DO
The only forces are gravity and the normal force.

∑F_y = NcosΘ - mg = 0
NcosΘ - mg = 0
NcosΘ = mg
N = mg/cosΘ

∑F_x = NsinΘ = mv²/r
NsinΘ = mv²/r
mg/cosΘsinΘ = mv²/r
tanΘ = v²/rg = 30² /(200*9.8)
Θ = 24.66381471 degrees

 

[Doc Al]

Your calculation is correct. Double check that you have the right info in the problem statement.

 

[bulbasaur88]

Thank you doc al :) I double checked and it is 27 degrees, but I am very certain that the answer key provided by my school is incorrect. Thank you :)

 

[PeterO]

Thank you doc al :) I double checked and it is 27 degrees, but I am very certain that the answer key provided by my school is incorrect. Thank you :)

To get 27 degrees you would have to use the sin function rather than the tan function at the very end.
I am very confident of the use of the tan function