Angle of average acceleration while turning a curve

In summary: Sorry, I had to fix a mistake. I hope I have it right now.In summary, the cyclist initially heading due North initiates a turn to the West with a quarter circle of radius 12-m and travels at a constant speed of 5.5-m/s. The direction of the average acceleration, measured anti-clockwise from East, is found by calculating the resultant of the initial and final velocity vectors and is equal to 225 degrees or 3.927 radians. This can also be calculated using calculus by taking the integral of the angle from 180 to 270 degrees divided by 90. The direction of average acceleration is therefore south west.
  • #1
sadpwner
8
0

Homework Statement


A cyclist is initially heading exactly due North. He then initiates a turn to the West, the turn being a quarter circle with radius 12-m. He travels at a constant speed of 5.5-m/s during the turn. What is the direction of this average acceleration, measured anti-clockwise from East?

Homework Equations


Theta=S/r

Where s is arc length.

The Attempt at a Solution


[/B]
0.25*2*12*Pi/12=1.57

This value seems way too small. Am I suppose to apply tan inverse to it which gives me 57 degrees?
 
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  • #2
Hi sadpwner:

I may be misinterpreting the problem statement, but it seems to me that "the direction of this average acceleration" equals the average of the direction of acceleration. If that is correct, then the actual speed is irrelevant, as long as it is a constant. What do you think?

Regards,
Buzz
 
  • #3
Buzz Bloom said:
Hi sadpwner:

I may be misinterpreting the problem statement, but it seems to me that "the direction of this average acceleration" equals the average of the direction of acceleration. If that is correct, then the actual speed is irrelevant, as long as it is a constant. What do you think?

Regards,
Buzz
I think the acceleration direction should be constant as the angle should be the same throughout the circle. I am not required to find the speed. I just need the direction of the acceleration in degrees.

I also just remembered that theta is measured in radians.

In that case, the working should be 360-1.57*180=77.4 degrees
 
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  • #4
sadpwner said:
0.25*2*12*Pi/12=1.57
I do not follow your calculation. What logic leads to that?
sadpwner said:
In that case, the working should be 360-1.57*180=77.4 degrees
That is not the right way to convert from radians to degrees.

What is the initial velocity as a vector? What about the final velocity? What vector do you need to add to the first to get the second?
 
  • #5
lol try 180*1.57?
 
  • #6
haruspex said:
I do not follow your calculation. What logic leads to that?

That is not the right way to convert from radians to degrees.

What is the initial velocity as a vector? What about the final velocity? What vector do you need to add to the first to get the second?
I just used theta=S/r formula.

Should be 180*1.57?

5.5m/s north initial. Then 5.5m/s west final. Using tan inverse it gives 45 degrees? However, that isn't the answer.
 
  • #7
sadpwner said:
I just used theta=S/r
That will give you the angle the cyclist turned through. That is not the same as the direction of acceleration.
When a planet goes in a circular orbit around its star, in which direction does it accelerate?
sadpwner said:
Should be 180*1.57?
No, that is not how you convert from radians to degrees. How many radians in a full circle? How many degrees in a full circle? What is the ratio between those two numbers?
sadpwner said:
5.5m/s north initial. Then 5.5m/s west final. Using tan inverse it gives 45 degrees? However, that isn't the answer.
Now you have calculated the bearing from the start of the arc to the end of the arc, NW. Again, that is not the acceleration.

Do you know how to find the resultant of two vectors?
Can you write the average acceleration vector in terms of the initial and final velocity vectors (and time)?
 
  • #9
The north component of speed decreases from 5.5 m /s to 0 m/s, so a southward (270 degree) component of acceleration. The west component of speed increases from 0 m/s to 5.5 m/s, so a westward (180 degree) component of acceleration. The direction of average acceleration is south west == 225 degrees == 3.927 radians. Or using calculus:

##\int_{180}^{270} \ \theta \ d\theta / 90 = 225##
 
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Related to Angle of average acceleration while turning a curve

1. What is the angle of average acceleration while turning a curve?

The angle of average acceleration while turning a curve refers to the average change in direction that an object experiences while moving along a curved path. It is a measure of how much the direction of the object's velocity changes over a given period of time.

2. How is the angle of average acceleration while turning a curve calculated?

The angle of average acceleration while turning a curve can be calculated by dividing the change in velocity (delta v) by the change in time (delta t) and then taking the inverse tangent of this value. This formula is represented as: θ = arctan(Δv/Δt).

3. What factors affect the angle of average acceleration while turning a curve?

The angle of average acceleration while turning a curve is affected by several factors, including the speed of the object, the radius of the curve, and the force applied to the object. Other factors such as the mass and friction of the object can also impact the angle of average acceleration.

4. Why is the angle of average acceleration while turning a curve important?

The angle of average acceleration while turning a curve is important because it is a measure of the object's change in direction, which can impact its overall velocity and motion. This information can be used to understand and predict the behavior of objects in curved motion, such as vehicles on a curved road or planets in orbit.

5. How does the angle of average acceleration while turning a curve relate to centripetal acceleration?

The angle of average acceleration while turning a curve is directly related to centripetal acceleration, which is the acceleration towards the center of a curved path. In fact, the angle of average acceleration can be calculated using the formula: θ = arccos(v^2/r*a), where v is the speed of the object, r is the radius of the curve, and a is the centripetal acceleration. This relationship helps to explain the forces at play in curved motion.

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