Solve Constrained Motion Question: Car Skids Off Path

[neemer]

[Mentors note: Thread moved from the Classical Physics forum after it had been replied to, hence the lack of a homework template]

Hey I have been trying to solve this problem for the last few days with no luck.

Question:

a car starts from rest at the origin and travels along the path given by y=0.2 x ^(3/2) and picking up speed in accordance with v = 4.8 t. At which x position does the car skid off the path? Friction force is known as well (Ff)Attempt at answer:

I parametrized the curve in terms of time, r(t) = <t, 0.2t^3/2> , calculated the radius of curvature (R) as a function of t, and used this to solve for the centripetal force using Fc = mv^2 / R , where v=4.8t . Then I said that the force of friction always acts in the inward normal direction to keep the car from going off the path. The road exerts a reaction force on the vehicle equal to - Fc . Therefore using Newtons second law i solved for the position t when Fc = Ff. This all makes sense to me although I've been told this is not the correct method. Any thoughts are appreciated. Thanks

 

[mfb]

y is some distance and v is speed? So we know y-distance as function of time and we know speed? Does that mean you have to find the x-position based on those two equations?

Or is it just a typo and the equation for y depends on x?

 

[neemer]

Ah yea my bad, the path is y(x) = 0.2 x^3/2 and the speed is a function of time

 

[Svein]

Since v(t) is not given as a vector, we must assume that it is the speed along the road. Now the arc length element ds is given by ds²=dx²+dy². Your equations gives y(x) = 0.2 x^3/2, so dy = 0.2⋅3⋅x²/2 dx or ds=\sqrt{1+0.3x^{2}}dx. Since v(t) = at, where a = 4.8m/s², the position at t = τ is given by s(\tau)=\frac{1}{2}a\tau^{2}. Now, the tangent vector \vec{t}= \frac{d\vec{r}}{ds} and the curvature is given as \kappa = \vert\frac{d\vec{t}}{ds}\vert. ...

 

[mfb]

Together with the centripetal force, we also have to take the acceleration along the direction of motion into account. The friction force has to "cover" both.

 

[neemer]

Thanks for the replies. Still a little confused.

The question is asking for the x position of the vehicle, not the distance covered along the path so would I still need to use arc length?

I think what I am getting confused about is direction of the friction force. Am I wrong to assume that the friction force only acts in the direction normal to the path of travel? Would there be a force that is always acting in the tangential direction Ft=4.8m since the velocity is increasing at a constant rate? Then the net force would be the tangential force plus the centripetal force and friction would have to act to oppose those combined forces?

Oh also note i corrected the equation of the path to make it more clear. Its y(x) = 0.2 x ^(3/2)

 

[mfb]

You'll need the arc length to find the relations y(t) and x(t), otherwise how would you know where you are at a specific speed?

Then the net force would be the tangential force plus the centripetal force and friction would have to act to oppose those combined forces?

Right.

 

[neemer]

Hmm okay. I have y(t) and x(t) now so I have r(t) = < x(t), y(t) > . Only thing is that these vectors are getting really complicated to the point i can only work with them in maple, and most of my class has never even seen vector calculus so I am starting to think there has to be a much simpler solution.

since the centripetal force is given by Fc= (mv^2) /r , and v is known, would it be possible to solve it using just this info? like when the magnitude of tangental and centripetal force components is equal to force of friction? So far I am just getting more confused.

 

[mfb]

You'll need r for that formula.
I think the curvature approach is a detour. x(t) and y(t) allow to calculate the acceleration in y and x direction, which gives the total acceleration.

 

[neemer]

I think i might have made a mistake getting x(t) and y(t).

I integrated ds from 0 to x so I had arc length as function of x, s(x) and then set s(x) = 1/2 *4.8*t^2 and solved for x. as far as I know this would be x(t) which i could use in my parametrization of the curve r(t) but this function is impossible to work it takes up a full line in maple and is full of square roots and such. Seeing as though we are not expected to know vector calc, there must be another solution method that I am missing. Even when i continue with maple to solve the problem I end get getting 4 solutions most of which are complex numbers.

 

[Svein]

Oh also note i corrected the equation of the path to make it more clear. Its y(x) = 0.2 x ^(3/2)

This changes several terms. Now dy = 0.2⋅(3/2)⋅x^(1/2)⋅dx, and ds=\sqrt{1+0.09x}dx...

 

[neemer]

Ah yea I realized that before and made the correction but thanks. What would x(t) be? I have an answer but its so long I am having a hard time thinking its right.

 

[Svein]

What would x(t) be?

Taking the indefinite integral on each side, we get s=\frac{2}{3\cdot 0.09}(\sqrt{1+0.09x})^{3}+C. SInce x=0 when s=0, C=0.

 

[neemer]

So then i set this equal to 1/2 * 4.8 ^ t^2 and solved so I have x(t). so now I have my position vector in terms of time. r(t)=<x(t) , 0.2*x(t)^(3/2) > , now differentiating this x2 i get a(t) i believe. Since the only force acting on the vehicle is friction, and this opposes the net acceleration, would the vehicle slip when the magnitude of a(t) = Ff ?

 

[mfb]

Looks right.

 

[Svein]

now differentiating this x2 i get a(t) i believe

Not quite. \frac{d\vec{r}}{dt}=\vec{v}(t)=(\frac{dx(t)}{dt},\frac{dy(t)}{dt}). The acceleration is then given by \frac{d\vec{v}}{dt}=\vec{a}(t)=(\frac{d^{2}x(t)}{dt^{2}}, \frac{d^{2}y(t)}{dt^{2}}). The magnitude of the acceleration is then √(a_x(t)²+(a_y(t)²).

 

[neemer]

Thanks for help, finally got the correct answer last night