Answer:
\[L=\,\,m\,\upsilon
\,r\] and \[\upsilon =r\omega =r\left( 2\pi n \right)\]
\[r=\frac{\upsilon
}{2\pi n}\] \[\therefore \] \[L=m\upsilon \left(
\frac{\upsilon }{2\pi n} \right)=\frac{m{{\upsilon }^{2}}}{2\pi n}\]
As \[K.E.\,=\frac{1}{2}m{{\upsilon
}^{2}}\], therefore, \[L=\frac{K.E.}{\pi n}\]
When
K.E. is halved and frequency (n) is doubled, \[L'=\frac{K.E.'}{\pi
n'}=\frac{K.E./2}{\pi \left( 2n \right)}=\frac{K.E.}{4\pi n}=\frac{L}{4}\]
i.e.
angular momentum becomes one fourth.
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