(a) Show that the position of a particle on a circle of...

(a) Show that the position of a particle on a circle of radius $R$ with its center at the origin is $\vec{r}=R(\cos \theta \hat{\imath}+\sin \theta \hat{\jmath})$, where $\theta$ is the angle the position vector makes with the $x$-axis. (b) If the particle moves with constant speed $v$ starting on the $x$-axis at $t=0$, find an expression for $\theta$ in terms of time $t$ and the period $T$ to complete a full circle. (c) Differentiate the position vector twice with respect to time to find the acceleration, and show that its magnitude is given by Equation $3.16$ and its direction is toward the center of the circle.

Key Concepts

Parametric Representation of Circular Motion

This concept involves describing the position of a point moving in a circle using trigonometric functions. The x and y coordinates are expressed as Rcos? and Rsin? respectively, providing a convenient way to relate the angular position to Cartesian coordinates.

Uniform Circular Motion

Uniform circular motion refers to motion along a circular path with constant speed. In this context, the constant speed implies that the angular position ? increases linearly with time, and the period T (the time to complete one full circle) relates directly to the angular velocity, with ? often expressed as 2?t/T.

Centripetal Acceleration and Differentiation in Motion

By differentiating the position vector with respect to time, one obtains the velocity and then the acceleration vectors. In circular motion, the second derivative yields a centripetal acceleration that always points toward the center of the circle, with a magnitude proportional to the square of the speed divided by the radius (or alternatively R times the square of the angular speed).

Transcript

00:01 In this exercise, we're asked to show first that the position vector for a particle going in a circular trajectory is given by r, cosine of theta i plus sine of theta j.

00:19 And in order to see that, let me draw here a coordinate system.

00:27 Okay.

00:28 In a circular trajectory and a certain and a vector pointing towards some point along this trajectory the vector r okay and i'm gonna zoom in here just so it's easier for us to work is the vector r this is the angle of theta okay and the magnitude of the vector r is capital r that's just the radius of the the circle.

01:08 Okay.

01:09 And we're going to project this vector onto the x and y axes.

01:15 So the projection of this vector into the y, the x x x is the magnitude of the vector capital r times the cosine of the angle that it makes with the x axis.

01:37 And the projection of the vector r in the y direction is given by the magnitude of the vector times sine theta j.

01:53 So this shows that the position vector is given by this formula here.

02:00 This is the answer to question a.

02:02 Now, on to question b, we are asked to find an expression for theta as a function of time, given that v, the speed of the particle is constant.

02:20 Okay.

02:23 And for that, we're going to first have to calculate the velocity vector v.

02:30 And in order to do that, we have to remember that v equals the derivative of r with respect to the time t.

02:38 Now, since we already know r, r is r cosine theta i plus r sine theta j, we can, we can differentiate this equation here.

03:01 And we're gonna have minus r sine theta.

03:07 And because of the chain rule, we have to differentiate what's inside the cosine.

03:12 So this is d theta d t plus r cosine cousine theta d theta d t.

03:25 And the first and second terms are our in the x and y directions.

03:33 And now we can calculate the speed, which is simply the magnitude of the velocity, which is v, okay, it's a constant, we know it.

03:45 And this is gonna be the square root of the sum of the components, the sum of the square of the component.

03:52 So it's gonna be r squared, sine square theta, d theta, d t squared, plus r squared cosine of theta squared d theta d t squared and this since sine squared plus cosine square equals one we're going to have square root of r squared d theta d t squared and this is r d theta d t okay, so we have an expression for d -theta d t that's v over r and if we integrate on both sides and take into account that the particle starts at t equals zero at theta equals zero.

04:56 We're going to have theta equals v over r times t.

05:02 Now, remember that v, the speed of the particle, is 2 pi r over the period of rotation tau.

05:15 I'm sorry, capital t.

05:19 So i can substitute this into the theta formula to get 2 pi t over top, over capital t.

05:31 Okay, that's the period of rotation.

05:33 And this is what the exercise asked us to do, to calculate theta as a function of t and the period capital t.

05:44 Okay...