Normal and tangential components of acceleration

AI Thread Summary
The discussion revolves around calculating the normal and tangential components of acceleration for a car transitioning from a straight path to circular motion. The car starts at rest with a tangential acceleration of 0.6 m/s², leading to confusion about the overall acceleration when entering the curve. Participants debate whether the tangential acceleration remains constant throughout the motion, impacting the calculations for speed at point B. The calculations suggest that if the tangential acceleration is constant, the speed at point B could be approximately 18.3 m/s, while the normal acceleration would be around 6.7 m/s². The clarity of the problem statement is questioned, as it affects the assumptions made about the car's acceleration during the turn.
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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?

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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.
 
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Using units, what did you get for its speed at point B? Please show us how you got this value.
 
Chestermiller said:
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
 
zachdr1 said:
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?
 
Show us your calculation of the distance between points A and B.
 
zachdr1 said:
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
 
Chestermiller said:
Show us your calculation of the distance between points A and B.
s_ab = 200+pi*50/4 = 239.27 m
 
zachdr1 said:
s_ab = 200+pi*50/4 = 239.27 m
That's not correct for either a clockwise or a counterclockwise path.
 
I think they meant for you to assume that the tangential component of acceleration is constant along the entire path.
 
  • #10
Chestermiller said:
That's not correct for either a clockwise or a counterclockwise path.
Sorry, I meant s_ab = 200 + 2pi(50)/4 =278.5m.
 
  • #11
Chestermiller said:
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
 
  • #12
zachdr1 said:
But then there'd be no normal component of acceleration
Who says?
 
  • #13
Chestermiller said:
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.
 
  • #14
zachdr1 said:
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.
 
  • #15
Chestermiller said:
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
 
  • #16
Sorry for diving in but..

zachdr1 said:
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.

zachdr1 said:
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.
 
  • #17
CWatters said:
Cars can usually accelerate around corners. At least they can as long as there is enough friction/traction. I'm with Chester.
Ditto.
 
  • #18
CWatters said:
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.
 
  • #19
zachdr1 said:
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.
 
  • #20
Chestermiller said:
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.
 
  • #21
Chestermiller said:
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.
 
  • #22
zachdr1 said:
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.
 
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