Simple Ice Skater with Conservation of Angular Momentum

[Lost1ne]

Homework Statement

Not a HW problem, but a "me re-thinking things" problem. Please tell me where my thinking is flawed:

You have an ice skater with no net external torques acting on him/her. (We are analyzing the time after they have to get an external torque on them by pushing off of the ground, and we are ignoring air resistance/friction slowing them back down.) Thus, their angular momentum is conserved. Thus, when they decrease their moment of inertia by moving their body in such a manner that more of their mass is closer to the axis of rotation (i.e., they pull their arms in towards their body), their angular velocity must increase, and vice versa.

But a change in angular velocity means that there must be an angular acceleration. If the ice skater then both a) has a moment of inertia about that axis of rotation, and b) is undergoing an angular acceleration of some magnitude, then, mathematically, there must be an external torque acting on them, right? But we're analyzing the problem under the assumption that the external torque is zero...I'm confused.

Homework Equations

Conservation of angular momentum of the ice skater about the vertical axis of rotation that runs through their center of mass. (L = I*w.) The time-derivative of angular momentum equaling zero as the net external torque on the ice-skater system equals zero.

 

[haruspex]

undergoing an angular acceleration of some magnitude, then, mathematically, there must be an external torque acting on them, right?

No, that's the point. A torque is required to change the angular momentum, but by reducing the moment of inertia it is possible to increase angular velocity without changing angular momentum.

 

[kuruman]

But a change in angular velocity means that there must be an angular acceleration.

Right.

If the ice skater then both a) has a moment of inertia about that axis of rotation, and b) is undergoing an angular acceleration of some magnitude, then, mathematically, there must be an external torque acting on them, right?

Wrong. The rate of change of angular momentum with respect to time is equal to the external torque, $\tau = \frac{dL}{dt}$. When there is no external torque but the moment of inertia changes, mathematically, $$0=\frac{dL}{dt}=\frac{d(I \omega)}{dt}=\omega\frac{dI}{dt}+I\frac{d\omega}{dt}=\omega\frac{dI}{dt}+I\alpha~ \rightarrow~\alpha =-\frac{\omega}{I}\frac{dI}{dt}$$ Thus, when the skater's moment of inertia is reduced as (s)he pulls in his/her arms, $dI/dt$ is negative, which makes $\alpha$ positive (same directon as $\omega$) which means that the angular speed increases.

 

[Lost1ne]

Right.

Wrong. The rate of change of angular momentum with respect to time is equal to the external torque, $\tau = \frac{dL}{dt}$. When there is no external torque but the moment of inertia changes, mathematically, $$0=\frac{dL}{dt}=\frac{d(I \omega)}{dt}=\omega\frac{dI}{dt}+I\frac{d\omega}{dt}=\omega\frac{dI}{dt}+I\alpha~ \rightarrow~\alpha =-\frac{\omega}{I}\frac{dI}{dt}$$ Thus, when the skater's moment of inertia is reduced as (s)he pulls in his/her arms, $dI/dt$ is negative, which makes $\alpha$ positive (same directon as $\omega$) which means that the angular speed increases.

Ohhhh, okay. From what I'm looking at, the equation of net torque = I*alpha is derived like you've shown from the time-derivative of angular momentum. However, t = I * alpha is derived under the assumption that the moment of inertia is constant, which is why that dI/dt term vanishes (it equals zero) and we obtain what you've shown.

In this sense, do we simply state that net torque = dL/dt is more fundamental than net torque = I * alpha?? I guess we would also say this analogously with net force and it being the time derivative of linear momentum versus the net force equaling mass * linear acceleration.

 

[haruspex]

net torque = dL/dt is more fundamental than net torque = I * alpha?

Yes!