A particle moves along a circular path over a horizontal x y...

A particle moves along a circular path over a horizontal $x y$ coordinate system, at constant speed. At time $t_{1}=5.00 \mathrm{~s}$, it is at point $(5.00 \mathrm{~m}, 6.00 \mathrm{~m})$ with velocity $(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}$ and acceleration in the positive $x$ direction. At time $t_{2}=10.0 \mathrm{~s}$, it has velocity $(-3.00$ $\mathrm{m} / \mathrm{s}) \hat{\mathrm{i}}$ and acceleration in the positive $y$ direction. What are the (a) $x$ and (b) $y$ coordinates of the center of the circular path if $t_{2}-t_{1}$ is less than one period?

Key Concepts

Uniform Circular Motion

Uniform circular motion describes the motion of an object traveling at constant speed along a circular path. In this type of motion, despite the constant speed, the object experiences a change in direction which produces an acceleration called centripetal acceleration. This acceleration is always directed toward the center of the circle, making the relationship between the velocity (which is tangent to the path) and acceleration (which is radial) a key characteristic of such motion.

Centripetal Acceleration

Centripetal acceleration is the acceleration required to keep an object moving in a circular path at constant speed. Its magnitude is given by v²/r, where v is the speed and r is the radius of the circle. This acceleration always points directly towards the circle’s center. The fact that the acceleration vector at any point is perpendicular to the velocity vector (which is tangent to the circle) is a crucial concept in analyzing the motion, as it allows one to determine the center of the circular path using geometric relations.

Geometrical Properties of Circles

The center of a circle has the property that it is equidistant from all points on the circle. In problems involving circular motion, one can use the perpendicularity between the position vector (from the center to the particle’s location) and the velocity vector (tangent to the circle) to derive relationships for the location of the center. In addition, the directions of the acceleration vectors, which point towards the center of the circle, play a vital role in determining the circle’s geometry from given points in time.

Relationship Between Velocity and Acceleration Directions

For an object in circular motion, the instantaneous velocity is tangential to the motion path, while the instantaneous acceleration (centripetal) is perpendicular to the velocity. Recognizing this perpendicularity allows one to infer that the radial (or centripetal) acceleration vector points along the line connecting the particle to the center of the circle. This fact is essential when using vector relations to determine the location of the center from known values of position, velocity, and acceleration at different times.

Transcript

00:01 This is chapter 4 problem number 66.

00:03 We're given a particle undergoing a uniform circular motion, and we're given it t equals one, pardon me, four seconds.

00:14 Velocity vector is 3j hat, and where the particle is at, at that point, the coordinates is 5 meters and 6 meters.

00:25 So let's assume that this is 5 meters, and we're here at 6 meters.

00:30 This is where the particle is at at t equals 4, t1, let's label it, and the velocity vector is 3j.

00:38 So it is in the positive y direction, right? this is v1.

00:45 And we're also given the acceleration, the information about the acceleration, which is in the positive x direction.

00:53 Again, this is our x and this is our y.

00:56 Then the projection of the circular motion looks like this, right? the acceleration is always towards the center of where the acceleration is pointing.

01:10 And we're also given at the bottom here the velocity vector pointing the negative extraction.

01:17 So v2 equals negative 3 i -hat when time t equals to 10.

01:27 Seconds.

01:29 And we're also given the acceleration.

01:31 So let me label the velocity vector first to the left, right, v2, then the acceleration vector is again towards the center.

01:40 Now, what we're asked is the x coordinate, x coordinate of the center of this rotation.

01:55 Given that, given enough time, is we have less time for the particle to go back to the initial state.

02:03 So basically how much distances traveled is three, one, two, three quarters of the circumference...

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