Normal and tangential components of acceleration

[zachdr1]

Homework Statement

A car starts at rest at point A, with a tangential component of acceleration a_t = 0.6m/s^2. Students approximate the acceleration of the car to be 0.6 m/s^2. What are the normal and tangential components of it's speed and acceleration at point B?

NOTE : the ends are supposed to be circular with radius 50, assume you're going CCW

Homework Equations

ads = vdv

The Attempt at a Solution

I tried this one in class for an hour and couldn't figure it out. I found the speed right when it gets to the circular motion to be 15.something, but I didn't think I had enough information to find anything else out.

 

[Chestermiller]

Using units, what did you get for its speed at point B? Please show us how you got this value.

 

[zachdr1]

Using units, what did you get for its speed at point B? Please show us how you got this value.

I didn't get the speed at point B. I didn't know how to get there. I got the speed at s=200 which was v=sqrt (2*a*s) = 15.49m/s

 

[gneill]

I didn't get the speed at point B. I didn't know how to get there. I got the speed at s=200 which was v=sqrt (2*a*s) = 15.49m/s

What assumption can you make about the linear (tangential) acceleration of the car? Do you think it will change when the car goes into the curve? Or, is the car's speed increasing steadily?

 

[Chestermiller]

Show us your calculation of the distance between points A and B.

 

[zachdr1]

I didn't get the speed at point B. I didn't know how to get there. I got the speed at s=200 which was v=sqrt (2*a*s) = 15.49m/s

the tangential acceleration has to decrease because there needs to be a normal component

but if you use 15.49^2/50 to find normal accel, that gives you 4.79 m/s^2 which is bigger than the total acceleration

 

[zachdr1]

Show us your calculation of the distance between points A and B.

s_ab = 200+pi*50/4 = 239.27 m

 

[Chestermiller]

s_ab = 200+pi*50/4 = 239.27 m

That's not correct for either a clockwise or a counterclockwise path.

 

[Chestermiller]

I think they meant for you to assume that the tangential component of acceleration is constant along the entire path.

 

[zachdr1]

That's not correct for either a clockwise or a counterclockwise path.

Sorry, I meant s_ab = 200 + 2pi(50)/4 =278.5m.

 

[zachdr1]

I think they meant for you to assume that the tangential component of acceleration is constant along the entire path.

But then there'd be no normal component of acceleration

 

[Chestermiller]

But then there'd be no normal component of acceleration

Who says?

 

[zachdr1]

Who says?

Because if the overall acceleration is 0.6, and the tangential acceleration is 0.6, then there has to be no normal component, which isn't possible

I would have understood if the problem didn't say that the overall acceleration was 0.6m/s^2, and if it said that the tangential acceleration was constant 0.6m/s^2 around the whole path. Then the speed would be 16.9 m/s at B.

 

[Chestermiller]

Because if the overall acceleration is 0.6, and the tangential acceleration is 0.6, then there has to be no normal component, which isn't possible

I would have understood if the problem didn't say that the overall acceleration was 0.6m/s^2, and if it said that the tangential acceleration was constant 0.6m/s^2 around the whole path. Then the speed would be 16.9 m/s at B.

See my post #9.

 

[zachdr1]

See my post #9.

They didn't state that though? And I asked my professor and he said that the acceleration is 0.6 for the entire motion.

Why even give the acceleration at A, after giving the tangential acceleration at A

 

[CWatters]

Sorry for diving in but..

I would have understood if the problem didn't say that the overall acceleration was 0.6m/s^2

It didn't actually say that.

And I asked my professor and he said that the acceleration is 0.6 for the entire motion.

and that's not 100% clear. He might have meant the car keeps accelerating at 0.6 regardless of the bend.

Cars can usually accelerate around corners. At least they can as long as there is enough friction/traction. I'm with Chester.

 

[gneill]

Cars can usually accelerate around corners. At least they can as long as there is enough friction/traction. I'm with Chester.

Ditto.

 

[zachdr1]

Sorry for diving in but..
It didn't actually say that.
and that's not 100% clear. He might have meant the car keeps accelerating at 0.6 regardless of the bend.

Cars can usually accelerate around corners. At least they can as long as there is enough friction/traction. I'm with Chester.

Damn you're right. I think the problem really just wasn't clear. This was a quiz problem, and what I assumed was that the car decreased in acceleration, and by the time it got to point B, it was at a constant speed of 5.6m/s or something like that. That was the only thing I could come up with under the assumption that he meant that the overall acceleration never exceeded 0.6m/s^2.

Oh well.

 

[Chestermiller]

They didn't state that though? And I asked my professor and he said that the acceleration is 0.6 for the entire motion.

Why even give the acceleration at A, after giving the tangential acceleration at A

The problem is much more complicated if you have to have the total acceleration at 0.6. But, it might turn out that the normal component of acceleration is very small compared to the tangential acceleration, and it might be possible to neglect it. To check this out, try a calculation in which you assume that the tangential acceleration is constant at 0.6 (rather than the overall acceleration). On this basis, calculate the velocity coming into the turn, and the velocity at point B. Then evaluate the radial component of acceleration at these locations, and see how they compare to the 0.6.

 

[zachdr1]

The problem is much more complicated if you have to have the total acceleration at 0.6. But, it might turn out that the normal component of acceleration is very small compared to the tangential acceleration, and it might be possible to neglect it. To check this out, try a calculation in which you assume that the tangential acceleration is constant at 0.6 (rather than the overall acceleration). On this basis, calculate the velocity coming into the turn, and the velocity at point B. Then evaluate the radial component of acceleration at these locations, and see how they compare to the 0.6.

I don't even think it's possible assuming that the total acceleration is 0.6. You end up not knowing either the tangential acceleration, velocity, or the time.

 

[zachdr1]

The problem is much more complicated if you have to have the total acceleration at 0.6. But, it might turn out that the normal component of acceleration is very small compared to the tangential acceleration, and it might be possible to neglect it. To check this out, try a calculation in which you assume that the tangential acceleration is constant at 0.6 (rather than the overall acceleration). On this basis, calculate the velocity coming into the turn, and the velocity at point B. Then evaluate the radial component of acceleration at these locations, and see how they compare to the 0.6.

Thank you for your help though. Hopefully I don't get too many points taken off that quiz.

 

[Chestermiller]

I don't even think it's possible assuming that the total acceleration is 0.6. You end up not knowing either the tangential acceleration, velocity, or the time.

At the beginning of the turn, the velocity is already $$v=\sqrt{(2)(200.)(0.6)}=15.5\ m/s$$ and the radial acceleration is already 4.8 m/s^2. So they must have wanted you to assume that the tangential acceleration is constant at 0.6. In that case, at point B, the tangential velocity would be $$v_B=\sqrt{(2)(278.5)(0.6)}=18.3\ m/s$$, and the radial acceleration would be 6.7 m/s^2.