A small puck on an air table revolves in a circle with rotational speed omega, h

Question

A small puck on an air table revolves in a circle with rotational speed omega, held at radius r by a weighted string that passes through a hole in the table. You slowly pull down on the weighted end of the string, decreasing the radius r. Assume that the angular momentum of the puck about the hole remains constant. What is the rotational speed when half the string has been taken up? The rotational speed has decreased by a factor of 2. The rotational speed has increased by a factor of 4. The rotational speed has increased by a factor of 2. The rotational speed hasnt changed. What has happened to the speed v during this time interval? The speed has increased by a factor of 2. The speed has increased by a factor of 4. The speed has decreased by a factor of 2. The speed hasnt changed.

Explanation / Answer

here,

Part B
L1 = L2
I1 * 1 = I2 * 2
I1 * v1/r1 = I2 * v2/r2
mr1^2 * v1/r1 = mr2^2 * v2/r2
v1r1 = v2r2
v2 = v1r1/r2

since r2 = (1/2)r1

v2 = 2v1

The speed has increased by a factor of 2

PART A:

as w1 = v/r

w2 = 2v/(1/2)

w2 = 4w1
The rotatinal speed has increased by a factor of 4

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