A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.90 s, it is at point (5.30 m, 5.30 m) with velocity (2.20 m/s) and acceleration in the positive x direction. At time t2 = 14.0 s, it has velocity (–2.20 m/s) and acceleration in the positive y direction. What are the (a)x and (b)y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit
To find the coordinates of the center of the circular path, we need to first find the positions of the particle at times t1 and t2, and then determine the midpoint of the line segment formed by connecting the two positions. Let's proceed step-by-step:
Step 1: Find the position of the particle at t1.
At t1, the particle is at point (5.30 m, 5.30 m). Therefore, the x-coordinate of the particle at t1 is 5.30 m.
Step 2: Find the position of the particle at t2.
At t2, we are given the velocity and acceleration of the particle. Since the acceleration is in the positive y direction and the velocity is in the negative x direction, we can conclude that the particle is undergoing centripetal acceleration, which always points towards the center of the circular path.
Since the particle is moving at a constant speed and undergoing centripetal acceleration, its path must be a circle. This implies that the displacement vector at t1 and t2, as well as the velocity vector at these times, are perpendicular to the radial line joining the center of the circle with the particle.
Therefore, we can draw a right-angled triangle with the given velocity vectors as its two legs, and the radial line connecting the two positions as its hypotenuse. The midpoint of the hypotenuse will be the center of the circle.
Using the given information, draw the triangle with the velocity vectors as legs:
(5.30 m, 5.30 m)
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|10.6\
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(unknown, unknown)
Step 3: Determine the midpoint of the line segment between the two positions.
The midpoint of the line segment between the two positions is the center of the circle. To find it, we need to find the midpoint coordinates using the given x-coordinate at t1 and the unknown y-coordinate.
Using the formula for finding the midpoint of a line segment, we can use the x-coordinate at t1 (5.30 m) and the distance between the two positions (10.6 m) to find the x-coordinate of the center:
x-coordinate of the center = (x-coordinate at t1 + distance between points) / 2
= (5.30 m + 10.6 m) / 2
= 7.95 m
So, the x-coordinate of the center is 7.95 m.
Step 4: Find the y-coordinate of the center.
Since the triangle formed by the velocity vectors and the radial line is a right-angled triangle, we can use the Pythagorean theorem to find the length of the radial line, which is also the distance between the two positions.
Using the given velocity magnitudes:
velocity at t1 = 2.20 m/s
velocity at t2 = 2.20 m/s
We can use the magnitudes as the lengths of the legs of the right-angled triangle:
velocity at t1 = 2.20 m/s
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velocity at t2 = -2.20 m/s
Using the Pythagorean theorem:
(distance between points)^2 = (velocity at t1)^2 + (velocity at t2)^2
= (2.20 m/s)^2 + (-2.20 m/s)^2
= 4.84 m^2 + 4.84 m^2
= 9.68 m^2
Solving for the distance between the points:
distance between points = sqrt(9.68 m^2)
= 3.11 m
Now, we can use the distance between the points and the y-coordinate at t1 to find the y-coordinate of the center:
y-coordinate of the center = (y-coordinate at t1 + distance between points) / 2
= (5.30 m + 3.11 m) / 2
= 4.205 m
So, the y-coordinate of the center is 4.205 m.
Therefore, the coordinates of the center of the circular path are (7.95 m, 4.205 m).
To find the x and y coordinates of the center of the circular path, we can use the fact that the velocity and acceleration vectors of the particle are directed towards the center of the circle at any given point on the circular path.
Let's start by finding the position vector of the particle at time t1. Since the particle is at point (5.30 m, 5.30 m) at t1, the position vector r1 is given by:
r1 = (5.30 m)i + (5.30 m)j
Next, we need to find the velocity vector and acceleration vector at time t1. The velocity vector v1 is given as (2.20 m/s), and since it is directed towards the center of the circle, the velocity vector is tangent to the circular path at this point. Therefore, v1 is perpendicular to r1. We can find v1 by taking the cross product of r1 and a unit vector in the positive z-direction.
v1 = r1 x k
where k is the unit vector in the positive z-direction.
Similarly, we can find the acceleration vector a1 at time t1. Since the acceleration is in the positive x-direction, it is directed towards the center of the circle. Therefore, a1 is also perpendicular to r1. We can find a1 by taking the cross product of r1 and a unit vector in the negative y-direction.
a1 = r1 x (-j)
Now, we repeat the same process to find the position, velocity, and acceleration vectors at time t2.
Let's find the position vector r2 first. Since the particle is at the same orbit, the magnitude of r2 will be the same as r1, but its direction will be different. We can find the direction of r2 by taking the dot product of r2 and r1.
dot(r1, r2) = |r1||r2|cos(phi)
where phi is the angle between r1 and r2. Since the particle is on the same orbit, phi = 0 degrees or 360 degrees.
dot(r1, r2) = |r1||r2|
|5.30||r2| = |5.30||r1|
|r2| = |r1| = 5.30 m
Therefore, the position vector r2 is given by:
r2 = (5.30 m)cos(phi)i + (5.30 m)sin(phi)j
Next, we find the velocity vector v2. Since v2 is directed towards the center of the circle and is perpendicular to r2, we can find v2 by taking the cross product of r2 and k.
v2 = r2 x k
Finally, we find the acceleration vector a2. Since a2 is in the positive y-direction, it is directed towards the center of the circle and is perpendicular to r2. We can find a2 by taking the cross product of r2 and (-j).
a2 = r2 x (-j)
Now that we have the position, velocity, and acceleration vectors at both t1 and t2, we can equate the magnitude and direction of the velocity and acceleration vectors to solve for the x and y coordinates of the center of the circular path.
For velocity:
|v1| = |v2|
For acceleration:
|a1| = |a2|
This will give us two equations to solve for the x and y coordinates of the center of the circular path.