You tape a piece of reflecting tape so that it runs from the center of a wheel t

Question

You tape a piece of reflecting tape so that it runs from the center of a wheel to its rim, so it looks like this:

You darken the room and use a camera to take a series of photos at times 0.050s apart. You trigger the camera so the first photo in the series (t = 0) occurs when the tape is horizontal, pointing to the right, at an angular displacement of ?0 = 0, and you take a series of five photos with the camera (at t = 0, 0.050 s, 0.100 s, 0.150 s, 0.200 s).

You darken the room and use a camera to take a series of photos at times 0.050s apart. You trigger the camera so the first photo in the series (t = 0) occurs when the tape is horizontal, pointing to the right, at an angular displacement of ?0 = 0, and you take a series of five photos with the camera (at t = 0, 0.050 s, 0.100 s, 0.150 s, 0.200 s).

For example, if the wheel didn’t rotate at all and stayed at ? = 0 for all times, your series of photos would look like this:

Explanation / Answer

a. 10 rev/s = 20pi rad/s

z= ang disp.

wo= initial angular velocity (which is constant)
z = wo*t
z = (20pi rad/s)*t
placing the different times we have
at t = 0 s, z = 0 rad
at t = 0.05 s, z = 1 rad
at t = 0.10 s, z = 2 rad
so the bar should move 1 rad every pic.

b. A = 25 rev/s^2 * 2pi rad/rev = 50pi rad/s^2
z = w0*t + (1/2)A*t^2
since w0 = 0 (starting from rest)
z = (1/2)A*t^2
z = (25pi rad/s^2)*t^2
substitute and you will get the positions

Z (0) = 0 , Z (0.05)= 0.0625rad, Z (0.1)=0.25, Z (0.15)= 0.5625, Z(0.2)= 1

c. w_0 = 10 rev/s = 20pi rad/s
A = -50 rev/s^2 = -100pi rad/s^2
z = w_0*t + (1/2)A*t^2
z = (20pi rad/s)*t + (-50pi rad/s^2)*t^2
just like in the past two substitute the values of t and you will obtain ang position.

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