Simple Rotation + Angular Velocity Question

[AiLove]

This problem is from Tipler's Physics for Scientists and Engineers, Chapter 9, Problem 15.
The tape in a standard VHS videotape cassette has a length L = 246 m; the tape plays for 2.0 h (Figure 9-36). As the tape starts, the full reel has an outer radius of about R = 45 mm, and an inner radius of about r = 12 mm. At some point during the play, both reels have the same angular speed. Calculate this angular speed in rad/s and rev/min.

I've thought about this problem for a while, and I don't understand what they're asking. I assumed the cassette was going at constant angular velocity, but I guess not. I looked at the answer for this problem, the first part of which is below, and it confused me even more. They seem to take some sort of average radius, and then use that as the radius for a w = vr equation. There is a diagram included, which just shows a VHS tape with the radii (12 mm and 45 mm) labeled.

1. At the instant both reels have the same area, 2(Rf^2 - r^2) = R^2 - r^2
2. Solve for Rf
Rf = 32.9 mm = 3.29 cm

Where is the [2(Rf^2 - r^2) = R^2 - r^2] equation from?

 

[Tide]

Those equations simply measure how much tape is on each reel. The left side indicates the two reels contain the same amount of tape (factor of 2) and equating that to the right side says that "tape is conserved." :)

 

[quicknote]

The equation [2(Rf^2 - r^2) = R^2 - r^2] is derived from the conservation of angular momentum. In this problem, the tape starts at rest, so the initial angular momentum is zero. As the tape plays, the outer reel spins faster and the inner reel spins slower, but the total angular momentum of the system remains constant. This means that the angular momentum of the outer reel must equal the angular momentum of the inner reel at some point during the play. This point is when both reels have the same angular speed, which is what the problem is asking for.

To solve for this angular speed, we can use the formula for angular momentum, L = Iw, where I is the moment of inertia and w is the angular velocity. The moment of inertia for a cylindrical object is given by I = 1/2mr^2, where m is the mass and r is the radius.

In this case, we can consider the outer reel as a point mass located at its average radius, Rf, and the inner reel as a point mass located at its average radius, r. This gives us the equation:

L = (1/2mRf^2)w = (1/2mr^2)w

We can then rearrange this equation to solve for w:

w = (mr^2)/(mRf^2)

Substituting in the values given in the problem, we get:

w = (m(0.012 m)^2)/(m(0.045 m)^2) = 0.711 rad/s

To convert to rev/min, we simply multiply by (60 s/2π rad) and get:

w = (0.711 rad/s)(60 s/2π rad) = 21.4 rev/min

Therefore, the angular speed when both reels have the same speed is approximately 0.711 rad/s or 21.4 rev/min. This calculation assumes that the tape is being played at a constant angular velocity, as you initially thought. However, the problem is asking for the angular speed at the point when both reels have the same speed, which is not necessarily the same as the average angular speed over the entire 2.0 hours. I hope this explanation helps clarify the solution for you.