Circular motion banked curve questions

[Samuelriesterer]

Relative equations:
F = ma
a = v^2/r
Fcent = mv^2/r
ωf^2 – ωi^2 = 2αs
s = θr
ω = v/r

Problem statement and work so far are all included:

(I am having trouble with 5 and no clue how to start on 7-9)

You are designing a roadway for a local business, to include a circular exit ramp from the highway. The curve has a radius of 30 meters. The exit ramp is designed to function properly with no friction holding the cars on the ramp. To accomplish this, the ramp must be banked or sloped from one side to the other.

1) If the design is to work for 1000 kg cars moving at 20 meters per second, calculate the force required to keep the car in its circular path.

Fcent. = mv^2/r = (1000 kg)*(20 m/s)^2/(30 m) = 13333.3 N

2) Draw a force diagram of the car on the banked turn. Indicate which side is the outer radius of the turn.

Only two forces, mg directed downward and N directed at an angle Θ left of the vertical axis.

3) Explain whether the normal force or the weight is larger and how you know (without calculation).

Normal force because it is the force that is moving the car.

4) Calculate the angle the ramp should be sloped.

Fnet_x = N sin θ = mv^2/r
Fnet_y = N cos θ = mg
N = mg/cos θ
mg/cos θ * sin θ = mv^2/r
mg *tan θ = mv^2/r
θ = arctan (20^2)/(30*9.8) = 53.68 deg

5) If the speed of a car on the ramp decreases from 20 m/s to 5 m/s after going around 3/4 of the 30 meter radius circle, determine the angular acceleration of the car.

ωf^2 – ωi^2 = 2αs
s = θr = (3π/2)*30
ωi = v/r = 20/30 = 2/3
ωf = v/r = 5/30 = 1/6
α = (ωf^2 – ωi^2)/2s = [(2/3)^2 – (1/6)^2]/2*(3π/2)*30 = < I know this isn’t right>

6) If the coefficients of friction between tires and the road are 0.6 and 0.3 (you figure out which is which), determine the maximum speed a 1000 kg car can have and still make it around the curve. Start by redrawing the force diagram, including friction.

N cos θ = mg + f sin θ
N = mg/(cos θ – μ sin θ)
Fnet = N sin θ + μ N cos θ = mg/(cos θ – μ sin θ) * (sin θ + μ cos θ) =
(mg tan θ + μ)/(1-μ tan θ) = mv^2/r →
v = sqrt [(rg tan θ + μ)/(1-μ tan θ)] = sqrt[(30*9.8 * tan 53.68 + .6)/(1-.6 * tan 53.68)] = 56.00 m/s

7) When the car is moving at 20 m/s, what is the speed of the center of the axle relative to the road?

?

8) Calculate the tangential speed (relative to the axle) of a point on the top of a 45 cm radius tire. explain your reasoning.

9) Calculate the tangential speed (relative to the road) of a point on the top of a 45 cm radius tire.

 

[haruspex]

3) Explain whether the normal force or the weight is larger and how you know (without calculation).

Normal force because it is the force that is moving the car.

Right answer, but the explanation could be a lot better. In what sense is the normal force 'moving' the car? Remember, forces accelerate.

ωf^2 – ωi^2 = 2αs

This equation is not even dimensionally correct. Think again.

(mg tan θ + μ)/(1-μ tan θ) = mv^2/r →

Something wrong with the bracketing there. Not sure if it was fixed up later.

7) When the car is moving at 20 m/s, what is the speed of the center of the axle relative to the road?

Well, what is the speed of the axle relative to the body of the car?

 

[Samuelriesterer]

3) the normal force is greater because it has a component in the direction of the acceleration. The force of gravity does not have any x component.

5) am I on the right track:

s = 3/2*pi*30 = 45pi m
a = (5^2-20^2)/(2*45pi) = -1.33 m/s^2
t = (5-20)/-1.33 = 11.28 s

angular acc = (3pi)/(2*11.28) = .42 rads/sec

I don't understan angular anything, at least how it is used in any function

7) the speed of the axle relative to the tire is zero, right. So the speed of the axle relative to the road is 20 m/s?

8) the tangential velocity of a rotating tire with respect to the center equals the forward speed of the tire, so:

v_tan = 20 m/s

9) ?

 

[haruspex]

3) the normal force is greater because it has a component in the direction of the acceleration. The force of gravity does not have any x component.

Almost there, but you also need to mention the vertical component of the normal force.

s = 3/2*pi*30 = 45pi m
a = (5^2-20^2)/(2*45pi) = -1.33 m/s^2
t = (5-20)/-1.33 = 11.28 s

Yes, that works (but please use brackets as appropriate), but the more direct route is to work with angles the same way you work with distances. If the angular acceleration is constant you can apply the usual SUVAT laws but with all distances replaced by angles:
$s \rightarrow \theta$, $v \rightarrow \omega$, $a \rightarrow \alpha$. So this equation:

ωf^2 – ωi^2 = 2αs

should have been ${\omega_f}^2-{\omega_i}^2 = 2\alpha\theta$

angular acc = (3pi)/(2*11.28) = .42 rads/sec

That's dividing an angle by a time. What does that give? (And do you mean (3pi/2)*11.28 ?)

So the speed of the axle relative to the road is 20 m/s?

Yes.

the tangential velocity of a rotating tire with respect to the center equals the forward speed of the tire, so:

v_tan = 20 m/s

Yes.

9) ?

You have just calculated the speed of the axle relative to the road, and the speed of the top of the tire relative to the axle. You now want the speed of the top of the tire relative to the road. How do relative velocities work?

 

[Samuelriesterer]

OK, so:

3)
Normal force because it is the force that is moving the car. There is an x component and a y component in the direction of acceleration. Conversely, the force of gravity only has a component in the y direction.

5)
ωf^2 – ωi^2 = 2αθ
θ = (3π/2)
ωi = v/r = 20/30 = 2/3
ωf = v/r = 5/30 = 1/6
α = (ωf^2 – ωi^2)/2θ = [(1/6)^2 – (2/3)^2]/2*(3π/2) = -3.93 m/s^2

9)
You add the velocities, so relative to the road the point's velocity would be 40 m/s?

 

[haruspex]

Normal force because it is the force that is moving the car.

Lose that bit - it doesn't mean anything. Objects can move without requiring a force to maintain the motion. Forces can accelerate objects, or balance other forces so as to prevent acceleration (positive or negative).

There is an x component and a y component in the direction of acceleration. Conversely, the force of gravity only has a component in the y direction.

Still not quite there. You need to say something quantitative. How does each component of the normal force compare to the corresponding component of the gravitational force?

5)
ωf^2 – ωi^2 = 2αθ
θ = (3π/2)
ωi = v/r = 20/30 = 2/3
ωf = v/r = 5/30 = 1/6
α = (ωf^2 – ωi^2)/2θ = [(1/6)^2 – (2/3)^2]/2*(3π/2) = -3.93 m/s^2

All correct except the units.

9)
You add the velocities, so relative to the road the point's velocity would be 40 m/s?

Yes.

 

[Samuelriesterer]

Thanks for your help.

I would say just mathematically I understand N is bigger because it equals mg divided by some fraction of 1.

 

[haruspex]

Thanks for your help.

I would say just mathematically I understand N is bigger because it equals mg divided by some fraction of 1.

Please try to answer my question:

How does each component of the normal force compare to the corresponding component of the gravitational force?

I.e.:

• what can you say about the relative magnitudes of the horizontal component of N and the horizontal component of the g force?
• what can you say about the relative magnitudes of the vertical component of N and the vertical component of the g force?