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?


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:

This equation has the limitation that the torque equation describes the instantaneous axis of rotation or center of mass for any type of motion – whether pure translation, pure rotation, or mixed motion. $I$ = Moment of inertia about the point which the torque is written (either instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then the torque equation is the same about all points in the plane of motion.

Few points I would like to emphasise on:

  1. In general, in the formula, $\mathbf{L} = I\boldsymbol{\omega}$, $I$ is the moment of inertia tensor and depending on the axis of rotation and our assumption of co-ordinate axes, it becomes a scalar or a vector and so on.
  2. In $\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$, $I$ is a constant, hence it is taken out of the term $\frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t}$ to form the term $I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}$.But here $I$ is not constant. So you cannot use that formula.
  • 1
    $\begingroup$ Top-notch explanation! $\endgroup$
    – rob
    Dec 16 '15 at 5:46
  • 1
    $\begingroup$ $\frac{dL}{dt}=I\frac{d\omega}{dt}+\omega \frac{dI}{dt}$ $\endgroup$
    – Floris
    Dec 16 '15 at 10:28

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