Inertia \(I\) of a discrete the mass \(m\) at distance \(r\) from an axis is given by (about

the axis)

\(I = mr ^2\)

\(\Rightarrow I \propto r^2\)

So, when \(r\) decreases inertia I decreases

Concept \(2\)

Angular momentum \((L)\) is a conserved quantity and remains constant when

external. Torque on a body is zero

\(\frac{ d \overrightarrow{ L }}{ dt }=\vec{\tau}\)

\(\text { if } \quad \tau=0\)

\(\Rightarrow \overrightarrow{ L }=\text { constant }\)

we know that

\(|\overrightarrow{ L }|= I . \omega \Rightarrow \omega=\frac{|\overrightarrow{ L }|}{ I }\) \(\Rightarrow \omega \propto \frac{1}{ I }\)

Now

when the dancer folds her arms basically whe is decreasing her inertia

And,

Since \(\omega \propto \frac{1}{I}\)

So, decreasing of inertia \((I)\), results, in inrease of angular velocity hence

The faster spin faster due to constant momentum and increase in angular

velocity