Inclined plane in a circular motion

[Mogdan]

Homework Statement

u=60km/hr
m=1400kg
Friction of road (wet road)=0.4
angle of depression=3.65 degrees
radius of circle=156.3m
Distance of total circumference=983m
Distance of circumference needed to go=154m
Therefore it must go 56.4/360 degrees to go that distance.
[/B]
My problem is to find the acceleration of a car going 60km/hr, with a mass of 1400kg, going down a hill on an angle of depression of 3.65 degrees. It is turning towards the right and the radius of the circle is 156.3m. I am confused on how to calculate this as I have come up with 2 formulas to use but I don't know how to use these formulas together to find the overall acceleration of the car coming down the hill.

For the first formula I don't understand by it is negative acceleration if the car is going down a hill. Would I just say it is accelerating at that value instead of decelerating.

Homework Equations

The first eformula is a=g(sin(theta)-Mu(cos(theta))
The second formula is a=(v^2/r) x (theta/360)

The Attempt at a Solution

For the acceleration on a declined plane I used the first formula and subbed in the variables known
a=9.8(sin(3.65)-0.4(cos(3.65))
a=-3.288m/s

Second formula is
a=v^2/r x (theta/360)
=60^2/156.3 x (56.4/360)
a=3.6m/s

 

[BvU]

Hello Mogdan,

My problem is to find the acceleration of a car going 60km/hr, down a hill

I think we first of all have to establish what the complete problem statement is. The way you put it, "zero m/s²" would be a suitable answer: the speed doesn't change in magnitude nor in direction.

Until it's turning to the right, of course. So do you want to calculate the acceleration at the start of that manoeuvre, or at some other point (or interval) in time ?

 

[Mogdan]

I just need to create a formula that would find the acceleration of a car that is on an incline while turning. This acceleration would then be used to sub into the formula for minimum braking distance (s=u^2/2 x mu x g)

 

[BvU]

Is this a homework exercise or are you making up all these values ?
And: is it clear to you that the acceleration you are referring to becomes a function of time if the direction on the incline changes ?

 

[Mogdan]

I will make it clear to you. Here is a picture of the road way the car will be coming down.

the orange line indicates where the vehicle will come from.
The red line indicates where the hill will be declined.
The car will be going down the hill at an angle of 3.65 degrees of depression while it is turning on a radius of 156.3m. How would I find the acceleration and/or the minimum braking distance if the car was traveling at 60 km/hr?

 

[haruspex]

find the acceleration and/or the minimum braking distance

The one does not translate directly into the other. As BvU posted, if the car is moving at a given speed then the slope does not affect the acceleration (the bend does). But the slope does affect the braking force available.

It is unclear whether the road is an arc on a sloping plane, in which case the slope will decrease as it tracks around the curve, or follows a helix, like a helter skelter, making the slope constant.

Finding the minimum braking distance is quite complicated. As the car slows down, the radial acceleration decreases, leaving more braking force available for slowing.

So please state the problem exactly:
- is it a sloping plane or a helix;
- do you need to find the minimum braking distance, the maximum initial deceleration, or... what?

 

[CWatters]

Quick back of the envelope calculation suggest the car should have no problem stopping. I estimated it should be able to stop in under a third of the available stopping distance.

 

[Mogdan]

It is in a helix. and I need a formula for which i will be able to plot in certain initial speeds of the car to find the minimum braking distance.

 

[Mogdan]

Quick back of the envelope calculation suggest the car should have no problem stopping. I estimated it should be able to stop in under a third of the available stopping distance.

You see I can't estimate how far the stopping distance will be. Once I find the minimum braking distance formula, I need to include in other factors such as driver reaction time, coefficient of the road and all sorts of other stuff to get an exact value. This assignment is way above the knowledge a year 11 should know. I don't know why they would give us something like this.

 

[haruspex]

You see I can't estimate how far the stopping distance will be. Once I find the minimum braking distance formula, I need to include in other factors such as driver reaction time, coefficient of the road and all sorts of other stuff to get an exact value. This assignment is way above the knowledge a year 11 should know. I don't know why they would give us something like this.

I suspect that whoever set the question did not appreciate its complexity.
If the slope is φ and the angle traveled around the curve at time t is θ then I get for max deceleration:
$\dot \theta^4+(\ddot\theta-\frac gr\sin(\phi))^2=\left(\frac gr\right)^2\mu^2\cos^2(\phi)$
where $\dot\theta$ is positive and $\ddot\theta$ is negative.
The only ways forwards would be a numerical approach (spreadsheet, say) or some simplifying bound.
E.g. you could, conservatively, take the centripetal force requirement as constant, as though it were doing max speed all along. That would give you an upper bound for the braking distance, but it might be rather too conservative.

Edit: it is interesting to note that if the initial speed is so great that $\dot\theta_0^2=\frac gr\sqrt{\mu^2\cos^2(\phi)-\sin^2(\phi)}$ then $\ddot\theta$ is zero, so no deceleration is possible. The tyres are doing all they can keeping the car from skidding off the road.

 

[Mogdan]

I suspect that whoever set the question did not appreciate its complexity.
If the slope is φ and the angle traveled around the curve at time t is θ then I get for max deceleration:
$\dot \theta^4+(\ddot\theta-\frac gr\sin(\phi))^2=\left(\frac gr\right)^2\mu^2\cos^2(\phi)$
where $\dot\theta$ is positive and $\ddot\theta$ is negative.
The only ways forwards would be a numerical approach (spreadsheet, say) or some simplifying bound.
E.g. you could, conservatively, take the centripetal force requirement as constant, as though it were doing max speed all along. That would give you an upper bound for the braking distance, but it might be rather too conservative.

Edit: it is interesting to note that if the initial speed is so great that $\dot\theta_0^2=\frac gr\sqrt{\mu^2\cos^2(\phi)-\sin^2(\phi)}$ then $\ddot\theta$ is zero, so no deceleration is possible. The tyres are doing all they can keeping the car from skidding off the road.

Yes I would need to do this in an excel spread sheet but this looks very complex, nothing like we have learned in class. Actually the teacher told us to teach ourselves on inlined planes and circular motion and centripetal acceleration. Would you be able to simplify this for me as I am having a hard time understanding? An example maybe?
Also when you say t is the time, would that be the time of the duration of the yellow lights which is 4 seconds?
How would I find the angle traveled around the curve? What do you mean by those thetas with a dot and subscript numbers with it. I have never seen them in my life.
When you also do cos^2(theta), how would i do that on my calculator?

Thanks haruspex for helping me with such a complex question.

 

[haruspex]

t is the time, would that be the time of the duration of the yellow lights which is 4 seconds?

t is the time from start of braking. What I posted is a differential equation describing what is happening at some arbitrary time t.

thetas with a dot and subscript numbers with it

$\dot\theta$ is the same as $\frac{d\theta}{dt}$, the rate of change of theta, i.e. the angular velocity. So the speed at time t is $r\dot\theta$.
Similarly, $\ddot\theta$ is $\frac{d^2\theta}{dt^2}$, the rate of change of angular velocity, i.e. the angular acceleration. So the tangential acceleration is $r\ddot\theta$.
$\dot\theta_0$ is the initial angular velocity, i.e. initial speed/radius.

cos^2(theta)

cos^2(phi), i.e. the square of the cosine of the slope angle.

Anyway, I did set it up on a spreadsheet and found that the curve is effectively irrelevant. The starting speed is so low in relation to the radius of the curve that the centripetal acceleration required to negotiate the bend is tiny compared with the braking available. So you can effectively treat it as a straight road.
The curve did not become important until I increased the starting speed to 80kph. On a straight road the braking time was then 6.8s, whereas on the curve it was 7.6s.
Increasing the speed beyond that very quickly made staying on the road impossible.

So I recommend that you
(1) calculate the centripetal force required initially
(2) calculate the braking force available
(3) argue that the second is so much greater than the first that the bend can be ignored

 

[Mogdan]

t is the time from start of braking. What I posted is a differential equation describing what is happening at some arbitrary time t.

$\dot\theta$ is the same as $\frac{d\theta}{dt}$, the rate of change of theta, i.e. the angular velocity. So the speed at time t is $r\dot\theta$.
Similarly, $\ddot\theta$ is $\frac{d^2\theta}{dt^2}$, the rate of change of angular velocity, i.e. the angular acceleration. So the tangential acceleration is $r\ddot\theta$.
$\dot\theta_0$ is the initial angular velocity, i.e. initial speed/radius.

cos^2(phi), i.e. the square of the cosine of the slope angle.

Anyway, I did set it up on a spreadsheet and found that the curve is effectively irrelevant. The starting speed is so low in relation to the radius of the curve that the centripetal acceleration required to negotiate the bend is tiny compared with the braking available. So you can effectively treat it as a straight road.
The curve did not become important until I increased the starting speed to 80kph. On a straight road the braking time was then 6.8s, whereas on the curve it was 7.6s.
Increasing the speed beyond that very quickly made staying on the road impossible.

So I recommend that you
(1) calculate the centripetal force required initially
(2) calculate the braking force available
(3) argue that the second is so much greater than the first that the bend can be ignored

Thanks for this help. Would you kindly be able to link me the excel spreadsheet; I would love to visually understand the graph.

 

[haruspex]

Thanks for this help. Would you kindly be able to link me the excel spreadsheet; I would love to visually understand the graph.

Uploads of spreadsheets don't seem to be allowed.

 

[Mogdan]

Uploads of spreadsheets don't seem to be allowed.

Maybe to an external site such as mediafire?

 

[Mogdan]

Could you at least upload some screenshots of it? Would be a great help.

 

[haruspex]

Could you at least upload some screenshots of it? Would be a great help.

See https://drive.google.com/file/d/0B4HHu7Lkl_FAb0paVXRkWDVxV3M/view?usp=sharing.
As I wrote, with the given numbers the bend is almost irrelevant, so the velocity-time graph would be a straight line. So for illustration in the spreadsheet I reduced the radius to less than half that given.
You can see how, at first, very little deceleration is possible. This is because almost all the available frictional force goes into providing the centripetal acceleration, leaving very little for tangential deceleration. As the speed reduces, more and more frictional force is available for braking, so the curve steepens and straightens.

 

[Mogdan]

Thanks so much for this.