Linear/tangential velocity, in a circlular path, increases with the increase in radius and decreases with the decrease in radius. Hence, the angular velocity remains the same no matter what the change in radius is(W=V/r). However, when we talk about the conservation of angular momentum, we say that since the momentum is conserved, as we increase the radius, the linear velocity must decrease to keep it constant(because L=mvr), which concludes it is inversely related to the radius, which concludes that angular velocity is inversely propotional to the square of radius, ( when an ice skater brings his arms inwards, it rotates with greater angular velocity)but It is clear from what's stated above, that angular velocity must remain the same, regardless of what the radius is, isn't it?
[Physics] Relation of Angular and Linear velocity with radius of circular path
angular momentumvelocity
Related Solutions
The asymmetric problem gets into more complicated aspects of the kinematics of rotating bodies than are usually the point when presenting this example.
In the initial, arms out symmetric configuration, the skater's center of mass is directly over the pivot point. If she brings one arm in, and still has the axis of her body strictly vertical, then her center of mass is no longer over the pivot point, and, were she not spinning, she would fall. Now, because she is spinning, you are dealing with a problem similar to that of a gyroscope in a gravitational field whose axis of rotation is non-vertical: the gyroscope precesses.
More step wise
1. (initial condition) the skater is spinning about a vertical axis, both arms outstretched.
2. skater starts pulling her left arm inward, this changes the location of her center of mass.
3. skater starts "falling" towards her outstretched right arm.
4. this is a torque (due to gravity and the friction on the ground that keeps her skate tip at a fixed point on the ice) on a spinning body.
5. this ends up causing her main axis of rotation to precess.
I believe that I can see these kinds of effects in some of the examples here. First thing to note is that the skate is always tracing a circle on the ice. The size of this circle is related to the degree of asymmetry in the skaters body position: more asymmetric - larger circle. This is consistent with the skater needing to manage the location of her center of mass by adjusting her body, in particular her legs, and/or needing to manange rotating about a non-vertical axis in such a way that her overall angular momentum is (very close to) exactly vertical.
This page has a nice summary of rigid body mechanics, which if worked through, could be applied to this situation.
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:
- 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.
- 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.
Best Answer
It seems to me that you mix up some things. What you are relating to is the angular momentum of a single point particle,
$$ \vec{l} = mrv\vec{e}_z = mr^2\omega \vec{e}_z, $$ which I have written here directly depending on the angular velocity $\omega$. So you see that a point particle has a fixed angular velocity $\omega$ as long as $l$, $r$ and $m$ are fixed. To change it, a torque has to be applied. Actually, it is defined as the change of angular momentum ($D = \dot{l}$).
When you are referring to the ice skater, it is important that this deals with the rotation of a rigid body, i.e. a system of individual point masses. The total angular momentum of this system is obtained by integrating (or for a discrete set of particles, summed) over all infinitesimal contributions of the point particles. This obviously depends on the geometry of the object. The information about the object is contained in the so-called tensor of inertia, $$ \vec{L} = I \cdot \vec{\omega} $$ To put a long story short, the ice-skater increases $\omega$ when contracting herself because on the system as a whole, there is no torque. Let's assume the ice-skater is a cylinder, for which one can find $I = \frac{m}{2}r^2$, so that $L = \frac{m}{2} r^2 \omega$. When the ice-skater is reducing her radius, we find a $\omega$-gain of $$ \frac{\omega_2}{\omega_1} = \frac{r_1^2}{r_2^2}. $$