Question

Flashlight

A 0.7- kg flashlight is swung at the end of a string in a horizontal circle of 0.45- m radius with a constant angular speed. If no torque is applied, what must the radius become if the angular speed of the flashlight is to be halved?

Answer 1

The angular momentum (which is the key here, since they're saying no torque is applied) must be conserved (or constant).

Angular Mom. = L = m*r² * w

Where m is mass, r is the radius, and w is the angular speed.

Now you know L1 = 0.7 * 0.45² * w

And you know L2 = 0.7 * r² * w/2

Meaning that your angular speed is halved, but mass of flashlight stays the same. Also, L1=L2 since no torques are used.

This gives:
0.7 *(0.45)² (w) = 0.7 (r)² (w/2)

or:

r² = 0.45² *2
or
r = 0.636396n