How Does Angular Velocity Change as String Length Reduces in a Rotating System?

[Simfish]

This is a problem from Kibble Classical Mechanics, so it may be harder than it looks.

Homework Statement

10. A particle of mass m is attached to the end of a light string of length
L. The other end of the string is passed through a small hole and is slowly pulled through it. Gravity is negligible. The particle is originally spinning round the hole with angular velocity ω. Find the angular velocity when the string length has been reduced to L/2. Find also the
tension in the string when its length is r, and verify that the increase in
kinetic energy is equal to the work done by the force pulling the string
through the hole.

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Attempt:

$$PE_{orig} = (1/2) k L^2$$

$$PE_{new} = (1/2) k (L/2)^2 = (1/8) k L^2$$

So $$PE_{new} / PE_{old}$$ is 4.

Now, I know that rotational kinetic energy is E = 1/2 I ω^2. And the answer says that the new angular velocity is 4 times that of the old angular velocity. But how do I get that?

Also, how do I get the tension in the string? Tension comes from a force opposing an applied force. Also, I know that W = Fd = F*L/2. So... maybe $$F = W/d = \Delta PE/d = (7/8) k L^2 / (L/2) = (7kL/16)$$. But that's still nowhere near the answer

 

[cupid.callin]

just use angular momentum conservation

 

[S.Mukherjee]

Let particle and string be a system
Since no external torques act, angular momentum is conserved.
Hence,