A flywheel is in the form of a solid circular | vedclass

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(B) The rotational kinetic energy $(K.E._{rot})$ is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^{2}$.First,calculate the moment of inerti

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(B) The rotational kinetic energy $(K.E._{rot})$ is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^{2}$.First,calculate the moment of inertia $(I)$ for a solid circular disc: $I = \frac{1}{2} M R^{2} = \frac{1}{2} 	imes 72 	imes (0.5)^{2} = 36 	imes 0.25 = 9 \ kg \cdot m^{2}$.Next,convert the angular velocity $(\omega)$ from $r.p.m.$ to $rad/s$: $\omega = \frac{70 	imes 2 \pi}{60} = \frac{7 \pi}{3} \approx 7.33 \ rad/s$.Now,substitute these values into the kinetic energy formula: $K.E._{rot} = \frac{1}{2} 	imes 9 	imes (7.33)^{2} \approx 0.5 	imes 9 	imes 53.72 \approx 241.7 \ J$.Rounding to the nearest provided option,the energy is approximately $240 \ J$.

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

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