Problem with motion in a circular path

In summary: I really appreciate it. In summary, the homework statement asks for the average vector velocity between 0 and pi/4 sec.
  • #1
8
0

Homework Statement


Calculate the average vector velocity between 0 and pi/4 sec.

Homework Equations


x=rcos2t
y=rsin2t
Vx=2rsin(2t)
Vy=2rcos2t
Ax=4rcos(2t)
Ay=-4cos(2t)
Circular path is x^2+y^2+r^2

The Attempt at a Solution


I'm not sure if I am missing something simple or not, What I need to know is what is r and so I initially set the square of the x and the square of the y equations equal to the square r. Pythagorean style. when I solved for r I felt just a little dumb because I got 1=1 which well duh I shoul've saw that coming. Now I'm trying to guess whether I even need r to do an average, or if there is another way to solve for r that I am missing. I also know that x=r when t=0 but when I sub r for x in the position equation I just get 1.
 
Physics news on Phys.org
  • #2
What are the x and y coordinates of the particle at time t = 0?
What are the x and y coordinates of the particle at time t = π/4?
What is the vector displacement of the particle (in terms of the unit vectors in the x and y directions) between time t = 0 and time t = π/4?
What is the definition of average (vector) velocity?

Chet
 
  • #3
At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.
 
  • #4
Heather said:
At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.
r is the radius of the circle, which is treated as an algebraic parameter.

chet
 
  • #5
ahh duh, ok so then it should look something like this, taking the root of the sum of the x and y coordinates for t=π/4 and the root of the sum of the x and y coordinates at t=0 subtract the final position from the initial, and divide by two. Giving (2-root2)/2.
 
  • #6
That's not correct. But, it's hard to tell what you did. Please, just answer, in order, the 4 questions I asked in post #2.

Chet
 
  • #7
Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).
 
  • #8
[
Heather said:
Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)

What is cos(2π/4) equal to?
What is sin(2π/4) equal to?
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0

Please express this in vector form using unit vectors or with an ordered pair of components.
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).

please express this in vector form using unit vectors or with an ordered pair of components.
What is the magnitude of this average velocity vector?

Chet
 
  • #9
#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial

#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi
 
  • #10
Heather said:
#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial

This should be written as
Initial position = r i
Final position = r j

Displacement vector = - r i + r j
#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi
What is ##\sqrt{r^2}## equal to?

They did not ask for the magnitude of the average velocity vector. They asked for the average velocity vector itself. This is equal to the displacement vector divided by the time interval pi/4. What do you get for that?

Chet
 
  • #11
The average velocity vector would be v=-(4r/π)i+(4r/π)j
 
  • #12
Heather said:
The average velocity vector would be v=-(4r/π)i+(4r/π)j
OK. We're done here. Nice job.

Chet
 
  • #13
yes! thank you so much for helping me through that problem, and for being so patient.
 

What is the definition of motion in a circular path?

Motion in a circular path refers to the movement of an object along a circular trajectory, where the object continuously changes its direction but maintains the same distance from a fixed point.

What causes an object to move in a circular path?

An object moves in a circular path due to the presence of a centripetal force, which acts towards the center of the circle and keeps the object in its circular trajectory.

How is the speed of an object in circular motion related to its acceleration?

The speed of an object in circular motion is directly proportional to its acceleration. This means that as the speed increases, the acceleration also increases, and vice versa.

What is the difference between uniform circular motion and non-uniform circular motion?

Uniform circular motion is when an object moves along a circular path at a constant speed, while non-uniform circular motion is when the speed of the object changes as it moves along the circular path.

Can an object in circular motion have a constant velocity?

Yes, an object in circular motion can have a constant velocity if the direction of its motion is constantly changing but the magnitude of its velocity remains the same.

Suggested for: Problem with motion in a circular path

Replies
3
Views
532
Replies
1
Views
1K
Replies
14
Views
4K
Replies
3
Views
961
Replies
4
Views
1K
Replies
2
Views
1K
Replies
16
Views
2K
Replies
1
Views
639
Back
Top