How Is Average Velocity Calculated in Circular Motion?

In summary, the bicyclist makes a three quarter turn of radius 1.5 m at constant speed. During the turn, the only horizontal force acting on the bicyclist is a static friction force of constant magnitude 250 N. The total mass of the bicycle and rider is 60 kg.
  • #1
Poetria
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Homework Statement



A bicyclist makes a three quarter turn of radius 1.5 m at constant speed. During the turn, the only horizontal force acting on the bicyclist is a static friction force of constant magnitude 250 N. The total mass of the bicycle and rider is 60 kg.

Homework Equations

The Attempt at a Solution



The path: http://www.mathgoodies.com/lessons/fractions/circles/circle_three_fourths_blue.gif

The circle's radius = 1.5
The bicyclist's speed: 2.5
Acceleration at the point A: (-25/6,0)
Velocity at the point A: (0, 2.5)
Velocity at the point B (the final point of the journey): (2.5,0)
And the average velocity (-0.53, -0.53)

I have computed everything correctly except for average velocity and I have no idea how to get the right angle. I guessed there should be minus signs (-1.25, -1.25) but of course magnitudes are wrong.
I realize that this is because of acceleration. You can't just reproduce strategies fit for linear motion, i.e. to divide the sum of final and initial velocity by two.


 
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  • #2
Look into the definition of average velocity. Hint: The velocities at the start and finish are irrelevant.
 
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  • #3
I already tried this. I understand that this is the displacement over time. So displacement in this case is simply 1/4 out of the circumference. I divided it by the time I computed (constant velocity is given).

Time = (3*((2*pi*1.5)/4))/2.5

Average velocity (vector legnth) = (((2*pi*1.5)/4))/2.82743 = 0.833334
 
  • #4
Poetria said:
I already tried this. I understand that this is the displacement over time. So displacement in this case is simply 1/4 out of the circumference. I divided it by the time I computed (constant velocity is given).

Time = (3*((2*pi*1.5)/4))/2.5

Average velocity (vector legnth) = (((2*pi*1.5)/4))/2.82743 = 0.833334
No. You seem to be trying to compute an average speed based upon the distance of the path. That's not average velocity.

Forget the length of the path. You want to start by finding the displacement vector: the vector that describes the final position with respect to the starting location. Do it in vector form. If you take the origin as the center of the track, then the initial position is <r,0>. What's the final position in this form?
 
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  • #5
Many thanks. :) I got it.

initial position: (1.5,0)
final position: (0, -1.5)

((1.5,0)+(0,-1.5))/2.82743
 
  • #6
Poetria said:
Many thanks. :) I got it.

initial position: (1.5,0)
final position: (0, -1.5)

((1.5,0)+(0,-1.5))/2.82743
Very close. But your calculation of the displacement is not correct. You want to take the final position and subtract the initial position to find the net displacement:

<displacement> = <final> - <initial>
 
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  • #7
Oh, right. One magnitude would be positive, which is wrong.

((0,-1.5)-(1.5,0))/2.82743

(-0.530517, -0.530517)
 
  • #8
this is decidedly a gap in my knowledge. I have to do more exercices.
 

Related to How Is Average Velocity Calculated in Circular Motion?

What is average velocity vector?

Average velocity vector is a mathematical concept used to describe the overall motion of an object. It takes into account both the magnitude and direction of an object's displacement over a specific period of time.

How is average velocity vector calculated?

Average velocity vector is calculated by dividing the total displacement of an object by the total amount of time it took to cover that displacement. The result is a vector quantity with both magnitude and direction.

What is the difference between average velocity vector and instantaneous velocity vector?

Average velocity vector describes the overall motion of an object over a specific period of time, while instantaneous velocity vector describes the motion of an object at a specific moment in time. Average velocity takes into account the object's entire displacement, while instantaneous velocity only considers the object's displacement at a single point.

Can average velocity vector be negative?

Yes, average velocity vector can be negative. This means that the object is moving in the opposite direction of its positive velocity. For example, if an object moves 10 meters north and then 5 meters south in a given time period, its average velocity vector would be -5 meters/second in the north direction.

How is average velocity vector useful in real-world applications?

Average velocity vector is a crucial concept in physics and engineering, as it helps us understand the motion of objects and how they change over time. It is used in fields such as mechanics, fluid dynamics, and electromagnetism to analyze and predict the behavior of moving objects. It is also used in everyday applications, such as calculating the average speed of a car during a trip or the average velocity of a runner during a race.

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