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 \(\omega\). Find the angular velocity when the string length has been reduced to \(\frac{1}{2} l\). 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.

The concept of conservation of angular momentum states that the initial angular momentum is equal to the final angular momentum when no external torque acts on the system. In this case, since gravity is negligible and only tension is acting, the angular momentum will be conserved. We can write the conservation of angular momentum equation as: $$L_i = L_f$$ For a rotating particle in a circle, the angular momentum can be written as: $$L = I\omega$$ Where, \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Now, let's find the moment of inertia:

The moment of inertia of the particle can be found using the formula: $$I = m r^2$$ So, the initial and final moments of inertia are: $$I_i = m l^2$$ $$I_f = m (\frac{1}{2}l)^2$$

Now that we have the moments of inertia, we can apply the conservation of angular momentum: $$I_i \omega_i = I_f \omega_f$$ Substitute the initial and final moments of inertia: $$m l^2 \omega = m (\frac{1}{2}l)^2 \omega_f$$ $$\omega_f = 4\omega$$ The angular velocity when the string length has been reduced to \(\frac{1}{2} l\) is \(4\omega\).

Centripetal force in the string is provided by the tension in the string, which can be found using the formula: $$F_c = T = m\omega^2 r$$ So the tension in the string when its length is r is \(T = m\omega^2 r\).

Kinetic energy of the particle can be found using the formula: $$K = \frac{1}{2} I\omega^2$$ Initial and final kinetic energies are: $$K_i = \frac{1}{2} m l^2 \omega^2$$ $$K_f = \frac{1}{2} m (\frac{1}{2}l)^2 (4\omega)^2$$ The increase in kinetic energy is: $$\Delta K = K_f - K_i = \frac{1}{2} m l^2 \omega^2 (3)$$ The work done by the force pulling the string can be found using the formula: $$W = \int F dx$$ Substitute the force and the limits of integration: $$W = \int_{l}^{\frac{1}{2}l} m\omega^2 r d r$$ Now, integrate the expression: $$W = m\omega^2 \left[ \frac{1}{2} r^2 \right]_{l}^{\frac{1}{2}l}$$ $$W = m\omega^2 \left[\frac{1}{2} \left(\frac{1}{2}l\right)^2 - \frac{1}{2} l^2\right]$$ $$W = \frac{1}{2} m l^2 \omega^2 (3)$$ We can see that the increase in kinetic energy \(\Delta K\) is equal to the work done by the force pulling the string \(W\), as required.