Conservation of Angular Momentum with
Decreasing Radius

## Question:

An object of 4 pounds mass rests on a smooth table a distance of
3 ft from the center. The object is attached to a cord which
passes down through a hole in the center of the table. Initially
the object is traveling around in a 3 ft radius circle with a 6
f/s tangential velocity. The cord is drawn down through the hole
in the table at a constant rate of 2 f/s, reducing the radius of
the object's path. Find the speed of the object at the
instant the radius of its path is 2 ft. Also find the amount of
work done in drawing the cord down to this point. Neglect
friction and the size of the object.
## Answer:

The angular momentum of this conservative system must remain
constant. Angular momentum is **r**x**p** where **r** is
the radius and **p** is the linear momentum. But **p** is
just the mass times the tangential velocity, so the initial
tangential velocity times the initial radius must equal the new
tangential veloctiy times the new radius. So the tangential
velocity at radius 2 ft is vt=6
f/s*(3/2)=9f/s. There is also a radial velocity of 2 f/s
so the speed is the magnitude of vector sum of those components
s=(81+4)^.5 = 9.22f/s.
The work done is the change in kinetic energy. First we get
the work to change the kinetic energy due to the change in
tangential velocity. The work, W, at any radius, r, is 1/2*m*v^{2}-1/2*m*vi^{2} where
v is the final tangential velocity and vi is the initial
tangential velocity. But v is vi*(ri/r) so
W=1/2*m*vi^{2}*((ri/r)^{2}-1). Plugging in
the numbers from this problem,
W=1/2*4/32.2*6^{2}((3/2)^{2}-1)=2.795
ft-lb. Next we get the work to change the kinetic energy
due to the radial velocity change from 0 to 2f/s. The additional
work would be 1/2*4*/32.2*2^{2}
or .248 ft-lb for a total of 3.04 ft-lb.

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