A person of mass m stands at the centre of a table moment of inertia I and rotating with angular velocity - If the person moves distance -a- from the center along the diameter- what will be the new angular velocity of rotating table-

The correct option is D Iω I+ma ^ 2 [ Iw=( I+m a^ 2 ) w^ '; w'= Iw I+m a^ 2 #160;; ...

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Byju's Answer

Standard XII

Physics

Superposition of Electric Fields

A person of m...

Question

A person of mass m stands at the centre of a table moment of inertia I and rotating with angular velocity $ω$ . If the person moves distance 'a' from the center along the diameter, what will be the new angular velocity of rotating table?

\frac{I ω}{{m a}^{2}}

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Remains same

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\frac{I ω}{{2 m a}^{2}}

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\frac{I ω}{{I + m a}^{2}}

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Solution

The correct option is D $\frac{I ω}{{I + m a}^{2}}$
$\begin{matrix} I w = (I + m a^{2}) w^{^{'}} w^{'} = \frac{I w}{I + m a^{2}} ∴ o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$

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A person of mass m stands at the centre of a table moment of inertia I and rotating with angular velocity ω. If the person moves distance 'a' from the center along the diameter, what will be the new angular velocity of rotating table?

The correct option is D IωI+ma2Iw=(I+ma2)w′w′=IwI+ma2∴optionDiscorrectanswer.

A person of mass m stands at the centre of a table moment of inertia I and rotating with angular velocity $ω$ . If the person moves distance 'a' from the center along the diameter, what will be the new angular velocity of rotating table?

The correct option is D $\frac{I ω}{{I + m a}^{2}}$
$\begin{matrix} I w = (I + m a^{2}) w^{^{'}} w^{'} = \frac{I w}{I + m a^{2}} ∴ o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$