Angle of average acceleration while turning a curve

AI Thread Summary
The discussion focuses on calculating the direction of average acceleration for a cyclist turning from North to West in a quarter-circle path. The cyclist maintains a constant speed of 5.5 m/s, and participants debate the correct approach to determine the average acceleration's direction. It is clarified that the average acceleration is influenced by changes in both the northward and westward components of velocity, leading to a resultant acceleration direction of 225 degrees (southwest). Misunderstandings about converting radians to degrees and the nature of centripetal acceleration are addressed, emphasizing the importance of vector analysis in the solution. The final conclusion confirms the average acceleration direction as southwest at 225 degrees.
sadpwner
Messages
8
Reaction score
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?
 
Physics news on Phys.org
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
 
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
 
Last edited:
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?
 
lol try 180*1.57?
 
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.
 
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)?
 
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##
 
Last edited:
Back
Top