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.

The angular momentum of this conservative system must remain constant. Angular momentum is rxp 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*v2-1/2*m*vi2 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*vi2*((ri/r)2-1). Plugging in the numbers from this problem, W=1/2*4/32.2*62((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*22 or .248 ft-lb for a total of 3.04 ft-lb.

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