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(D) Given,mass of a circular disc,$M = 12 \,kg$,radius,$R = 0.5 \,m$,angular velocity,$\omega = 100 \,rad/s$.The moment of inertia of a thin circular

Vedclass · Accepted Answer

$7.5$

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(D) Given,mass of a circular disc,$M = 12 \,kg$,radius,$R = 0.5 \,m$,angular velocity,$\omega = 100 \,rad/s$.The moment of inertia of a thin circular disc about its central axis is $I = \frac{1}{2} MR^2$.The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.Substituting the value of $I$ in the formula,we get $K = \frac{1}{2} (\frac{1}{2} MR^2) \omega^2 = \frac{1}{4} MR^2 \omega^2$.Now,substituting the given values:$K = \frac{1}{4} 	imes 12 \,kg 	imes (0.5 \,m)^2 	imes (100 \,rad/s)^2$$K = 3 	imes 0.25 	imes 10000 \,J$$K = 0.75 	imes 10000 \,J = 7500 \,J$.Since $1 \,kJ = 1000 \,J$,we have $K = 7.5 \,kJ$.Therefore,the rotational kinetic energy of the disc is $7.5 \,kJ$.

$A$ thin circular disc of mass $12 \,kg$ and radius $0.5 \,m$ rotates with an angular velocity of $100 \,rad/s$. The rotational kinetic energy of the disc is (in $\,kJ$)

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