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### Circular Motion in a Cassette Player

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Date: 11/7/95 at 7:1:17
From: Anonymous
Subject: Circular motion problem

Dear Drs. Math,

I have been wrestling with a problem set to my A level class.
I have a real block on it and keep getting stuck.

The tape in a cassette player passes over the head at 5 cm per second
onto a takeup spool of 5 cm diameter.  The tape is 0.01 cm thick.
Find an expression for the angular speed of the takeup spool at time
t seconds.  What modelling assumptions do you make?

I would assume that the tape is taken up smoothly so the radius of
rotation does not increase in a sharp jerk in each turn.  I can derive
the angular speed at the end of a particular turn and time but I can't
get it all to come together into a neat expression linking angular
velocity and time.
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Date: 11/7/95 at 22:10:24
From: Doctor Jonathan
Subject: Re: Circular motion problem

Unless I'm solving this a dumb way, I end up getting a first order
non-linear differential equation - fortunately, one that is easy to
solve.

For what it's worth, here's what I did:

You know that the angular velocity will be the tangential speed (i.e.
the tape speed divided by the radius of the spool). So

w = 5/r

With every turn of the spool, the radius will increase by .01.  The
angular velocity divided by 2Pi is the rotations per second, so

dr/dt = .01w/2Pi

If we substitute the first expression for w into this equation, we end
up with

dr/dt = .01*5/2Pi / r = .008/r

r'*r = .008

The derivative of r needs to cancel out all factors of t in r, so the
only thing that works is r(t) = a*r^.5 + b where a is any constant, and
b is an initial condition for r.

We know that at time t = 0 r is 2.5, so b = 2.5. If we plug our expression
for r(t) into r'*r = .008, we find that a = 0.126. Thus

r(t) = 0.126*Sqrt[r] + 2.5

Plugging this into w = 5/r, we arrive at the following expression for w:

w = 5/(0.126*Sqrt[t] + 2.5)

This function yields a correct initial condition for w, and also
decreases over all positive t.  This and the fact that the decrease
becomes less steep as time goes on (the tape is having less effect on
the spool) leads me to believe that this is correct.

I really hope I have answered the question you wanted me to.

- Doctor Jonathan,  The Geometry Forum
```

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Date: Thu, 19 Sep 1996 09:49:22 -0400
From: Rick Brown
Subject: comments on an old problem

Dear Doctor Jonathan:

I've just discovered your Math Forum website, and I've been having a
lot of fun exploring it.  While perusing the "Ask Dr. Math" archives,
I found a problem called "Circular Motion in a Cassette Player," which
beyond" section).  The answer you gave involved a first order
non-linear differential equation.

Well, since I know how much Dr. Math enjoys slapping his forehead, I
thought I'd suggest a couple of things about your solution: (1) It's
slightly in error (apparently due to an algebraic slip when you were
solving the diff. eq.); and, much more importantly, (2) it can be
solved using simple geometry, and doesn't require any calculus at all!

First, let's address item (1).  You had set up a differential equation
which looked something like this:

r'(t) * r(t) = .008     (eq. 1)

You stated the general solution for this as being "r(t) = a*r^.5 + b".
Now, I think that second "r" was a typo and you meant to say "r(t) =
a*t^.5 + b", but in any case that solution is incorrect; it yields
this:

r'(t) * r(t) = 0.5*(a^2 + a*b*t^.5)

which, being a function of t, can't equal the constant .008.  I
believe the correct general solution is:

r(t) = a*(t + b)^.5     (eq. 2)

This yields: r'(t) * r(t) = (a^2) / 2, which, with (eq. 1) gives a =
.126.  Next, setting our initial condition r(0) = 2.5, and combining
that with (eq. 2), we get:

a*(0 + b)^.5 = 2.5

or: b = (2.5/a)^2 = (2.5/.126)^2 = 393.7.  So the solution for r is:

r(t) = 0.126 * Sqrt[t + 393.7]     (eq. 3)

Okay, now for the fun stuff.  Let's forget about all this differential
equation business, and find r(t) by geometry.

We know the length L of tape which has been taken up by the reel, as a
function of t.  This is just:

L(t) = v * t     (eq. 4)

where v = 5 cm/sec, the speed of the tape through the tape head.  Now
observe that the mass of tape which is on the reel at any particular
time forms a particular geometric shape: namely, a cylinder of radius
r, minus a smaller concentric cylinder of radius r0 = 2.5 cm.  (Note:
this is an approximation which relies on the assumption that the
thickness of the tape is quite small compared with r0; however, this
assumption was in fact made in the original statement of the problem
("...the radius of rotation does not increase in a sharp jerk in each
turn.").)

We can easily calculate the volume of this cylinder-with-a-hole, given
the height h of the tape.  It is:

Volume(t) = pi * (r(t)^2 - r0^2) * h     (eq. 5)

Now here's the crucial observation:  Given the thickness z of the
tape, we can also express its volume as though it were a long, thin
"box", like this:

Volume(t) = L(t) * z * h     (eq. 6)

So, combining equations 4, 5 and 6, we get:

v * t * z * h = pi * (r(t)^2 - r0^2) * h

or:

v * t * z = pi * (r(t)^2 - r0^2)

or:

r(t) = Sqrt[v * t * z / pi + r0^2]

Substituting v = 5 cm/sec, z = 0.01 cm, and r0 = 2.5 cm gives:

r(t) = Sqrt[(.01592 cm^2/sec) * t + 6.25 cm^2]

With a little rearrangement (and accounting for rounding error), this
can be seen to be equivalent to (eq. 3).

The original problem asked for the angular speed w(t) of the takeup
reel.  This, as you recognized, is just v/r(t) or, in its most general
form:

w(t) = v / Sqrt[v * t * z / pi + r0^2]

Sincerely,
Rick Brown
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Associated Topics:
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