12). A spoked wheel of mass M = 3 kg and radius R = 2 m is...

12). A spoked wheel of mass $M = 3 \text{ kg}$ and radius $R = 2 \text{ m}$ is free to rotate about a horizontal frictionless axel as shown at right. The free end of a massless rope wrapped tightly around the wheel is use to apply a constant tangential force of magnitude $F_A = 10 \text{ N}$ as shown. As the rope unwinds, the wheel undergoes a constant angular acceleration. If the wheel can be treated as a ring, the magnitude of the angular acceleration is: A). $2.04 \text{ rad/}s^2$ B). $1.04 \text{ rad/}s^2$ C). $0.6 \text{ rad/}s^2$ D). $1.67 \text{ rad/}s^2$

Transcript

00:01 So in this question we are given a string wrapped around a bicycle wheel that is mounted such that it can spin freely.

00:08 And when a force f is applied, let me draw, when a force f is applied horizontally to the string, the wheel begins to rotate and after our string has moved through a distance d, our bicycle wheel is, so we're going to call this our distance d, when our bicycle wheel, or sorry, when our string has moved through a distance d, our bicycle wheel has some angular speed and we are going to find that angular speed in terms of the moment of inertia of the wheel and the applied force and this displacement.

00:58 So one of the first things to notice is that when my, i drew this with some specific points for a reason.

01:08 So the point of application or the point of contact of the, excuse me, of the string with the wheel when we're pulling horizontally is right here at the top of this vertical spoke of the axis or a vertical spoke of the wheel.

01:24 And that's the point that i chose to move a distance d because what's going to happen is this other point that i drew 90 degrees away just because it was easy to draw is going to have moved also a distance d around the circumference of the circle in that time.

01:41 And likewise, if you take any point on the rim of the bicycle, it will also have moved a distance d.

01:50 And this of course is going to be an angle, i'm going to call it delta theta, that our wheel moves through in that time.

02:01 It doesn't have to be 90 degrees.

02:03 Again, i just chose 90 degrees to make this easy to solve.

02:07 So our key relationship here is going to be the arc length, which you might remember from geometry, arc length s equals the radius times our angular displacement.

02:18 Well, in this case, our arc length is d, which will be equal to r times the angular displacement.

02:29 So our angular displacement of our wheel over this time t will be equal to d divided by r...

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