Question

A disk of radius R = 2:0 cm rotates at a constant angular velocity of w = 8:165 rad/sec. Let us imagine that we are interested in the xy coordinates of a particular point on the perimeter of the disk. Furthermore, suppose that, at t = 0; our point of interest is exactly at the location (x,y)=(R, 0). The origin of the coordinate system is at the center of the disk.

(a) Write down an expression for the x coordinate x(t) as a function of time, and identify the analogous quantities amplitude, frequency, and the initial phase, for the rotating disk.

(b) Differentiate your expression for x(t) from part (a) to obtain an expression for v(t); and show that the amplitude of the velocity in the x direction is Rw; which just happens to be the tangential velocity for a point moving in a circle. Think about why this must be the case.

(c) Differentiate your expression for v(t) from part (b) to obtain an expression for a(t); and show that the amplitude for the acceleration in the x direction is Rw^2; which you should recognize to be the centripetal acceleration if you set w = v/R. Think about why this makes sense.

(d) Show that your expressions for x; v;and a; are identical to those found in problem 6; in spite of the fact that the rotating disk and the oscillator are two different physical systems, provided that you identify the radius R for the disk with the amplitude A for the oscillator.

Answer 1

(a)

Step 1) Using trigonometry, the x-coordinate for a circle of radius R centered at the origin can be written in terms of cosine. Since , the x-coordinate can be written as . The angle varies with time though. By definition, angular speed is equal to the change in the angle over the change in time t, or . Rearranging, . Plugging into , we get that the equation for the x-coordinate is . Now notice we need to be careful that there may be an initial angle that the point on the disk starts at. Since angles are measured from the positive x-axis, and we start at the positive x-axis, the initial phase is , the equation for the x-coordinate is .

Step 2) Plug in the given values for and into .

The equation for the x-coordinate of the point on the disk is . Be careful to note that if you'd prefer to use sine, you can use the trigonometric identity . If using the sine function instead, the initial phase isn't 0, it is so:

Step 3) The amplitude A of the position function x(t) occurs when the cosine function is equal to its max, which is 1. The amplitude of x(t) is thus equal to the radius of the disk, so .

Step 4) The frequency is given by . Since , the frequency is .

Step 5) As stated in step 1, the initial phase is 0, or .

(b)

Step 6) To find the velocity function v(t), differentiate the position function with respect to time.

The velocity function is given by .

Step 7) The amplitude (maximum) of the velocity occurs when the sine function reachese its max of 1 or -1. This means the maximum velocity is equal to , which is the radius times the angular speed or . Notice that this is just the normal equation relating the tangential and angular speeds of any rotating object, .

(c)

Step 8) Next differentiate the velocity function with respect to time to get the acceleration function.

Step 9) The amplitude (maximum) of the acceleration occurs when the cosine function reaches its max of 1 or -1. This means the maximum acceleration is equal to , which is the radius times the angular speed squared or . Centripetal acceleration has the equation . Since , the centripetal acceleration is , which is the same as what we got.

(d) Problem #6 wasn't given to compare this problem with.