Question

a fan of moment of inertia 0.6kgm^2 is to run upto a working speed of 0.5 revolution per secound correct angular momentum​

Answer 1

Explanation:Angular momentum, J= I ω

J= (0.6 × 2 π r ) Kg. m 2 s −1

= 0.6 × π ( 2 × 0.5 ) kg. m 2 s − 1

=0.6π kg.m2s-1

Answer 2

The angular momentum of the fan is $1.884\ kg-m^2/s$.

Explanation:

It is given that,

Moment of inertia of the fan, $I=0.6\ kg-m^2$

Angular velocity of the fan, $\omega=0.5\ rev/s$

or $\omega=3.14\ rad/s$

Let L is the angular momentum of the fan. The angular momentum in terms of moment of inertia is given by :

$L=I\omega$

$L=0.6\times 3.14$

$L=1.884\ kg-m^2/s$

So, the angular momentum of the fan is $1.884\ kg-m^2/s$. Hence, this is the required solution.

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Moment of inertia

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