An ice skater is spinning in a circle with an angular velocity, $\omega$, and rotational inertia $I$. Let's say that the skater then pulls his arms directly inwards at a constant rate over a time of $t$ seconds, so that his rotational inertia is halved to $\frac{I}{2}$. There are forces from the skater's muscles going through his center, but the lever arm is 0 so net-torque is 0. By conservation of angular momentum, his angular velocity would be doubled to $2\omega$.
Another equation states that $\tau = I\alpha$. Since angular velocity is changing, there is surely an angular acceleration. The skater also definitely has a non-zero rotational inertia, which implies there is a net-torque on the system.
This is a contradiction. Where have I gone wrong in my reasoning?
Best Answer
As you can see from this wiki link, the relation between torque and angular acceleration is derived in this way:
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
$$\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$ where $L$ is the angular momentum vector and $t$ is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
$$\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$ For rotation about a fixed axis,
$$\mathbf{L} = I\boldsymbol{\omega}$$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. It follows that
$$\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$$ where $α$ is the angular acceleration of the body, measured in $rad/s^2$.
Your answer specifically lies in the following paragraph:
Few points I would like to emphasise on: