Circular motion banked curve questions

In summary: I don't know how to represent with only 1 value7) the speed of the center of the axle relative to the road is 20 m/s8) the tangential velocity of a rotating tire with respect to the center equals the forward speed of the tire, so:v_tan = 20 m/s9) the speed of the top of the tire relative to the road is 20 m/s
  • #1
Samuelriesterer
110
0
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.
 
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  • #2
Samuelriesterer said:
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.
Samuelriesterer said:
ωf^2 – ωi^2 = 2αs
This equation is not even dimensionally correct. Think again.
Samuelriesterer said:
(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.
Samuelriesterer said:
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?
 
  • #3
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) ?
 
  • #4
Samuelriesterer said:
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.
Samuelriesterer said:
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:
Samuelriesterer said:
ωf^2 – ωi^2 = 2αs
should have been ##{\omega_f}^2-{\omega_i}^2 = 2\alpha\theta##
Samuelriesterer said:
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 ?)
Samuelriesterer said:
So the speed of the axle relative to the road is 20 m/s?
Yes.
Samuelriesterer said:
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.
Samuelriesterer said:
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?
 
  • #5
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?
 
  • #6
Samuelriesterer said:
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).
Samuelriesterer said:
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?
Samuelriesterer said:
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.
Samuelriesterer said:
9)
You add the velocities, so relative to the road the point's velocity would be 40 m/s?
Yes.
 
  • #7
Thanks for your help.

I would say just mathematically I understand N is bigger because it equals mg divided by some fraction of 1.
 
  • #8
Samuelriesterer said:
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:
haruspex said:
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?
 

Related to Circular motion banked curve questions

1. What is circular motion on a banked curve?

Circular motion on a banked curve is when an object moves in a circular path while also being inclined at an angle. This type of motion is often seen in roller coasters or race cars on a curved track.

2. What causes an object to bank in circular motion?

The centripetal force, which is directed towards the center of the circular path, causes an object to bank in circular motion. This force is necessary to keep the object moving in a curved path instead of flying off in a straight line.

3. How does the angle of the banked curve affect circular motion?

The angle of the banked curve affects the amount of centripetal force required to keep the object moving in a circular path. A steeper angle requires a greater centripetal force, while a shallower angle requires a smaller centripetal force.

4. What is the relationship between speed and banked curve angle in circular motion?

The relationship between speed and banked curve angle is known as the banking angle equation. It states that as the speed of the object increases, the angle of the banked curve must also increase in order to maintain a constant centripetal force and keep the object on its circular path.

5. How is the banked curve angle calculated in circular motion?

The banked curve angle can be calculated using the banking angle equation: tanθ = (v^2 / rg), where θ is the banked angle, v is the speed of the object, r is the radius of the circular path, and g is the acceleration due to gravity. This equation can be rearranged to solve for any of the variables.

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