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(B) The rotational kinetic energy $(K.E._{rot})$ of a rotating body is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^2$. For a solid circul

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(B) The rotational kinetic energy $(K.E._{rot})$ of a rotating body is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^2$. For a solid circular disc,the moment of inertia $(I)$ about its central axis is $I = \frac{1}{2} M R^2$. Given: $M = 72 \, kg$,$R = 0.5 \, m$. $I = \frac{1}{2} 	imes 72 	imes (0.5)^2 = 36 	imes 0.25 = 9 \, kg \cdot m^2$. The angular velocity $(\omega)$ in $rad/s$ is calculated from $r.p.m.$ as: $\omega = \frac{2 \pi N}{60} = \frac{2 	imes \pi 	imes 70}{60} = \frac{7 \pi}{3} \, rad/s$. Substituting these values into the kinetic energy formula: $K.E._{rot} = \frac{1}{2} 	imes 9 	imes (\frac{7 \pi}{3})^2 = \frac{1}{2} 	imes 9 	imes \frac{49 \pi^2}{9} = \frac{49 \pi^2}{2}$. Using $\pi^2 \approx 9.87$,$K.E._{rot} \approx \frac{49 	imes 9.87}{2} \approx 241.8 \, J$. Rounding to the nearest given option,the value is approximately $240 \, J$.

$A$ flywheel is in the form of a solid circular disc of mass $72 \, kg$ and radius $0.5 \, m$. If it rotates at $70 \, r.p.m.$,then the rotational kinetic energy is approximately ....... $J$.

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