A thin uniform circular disc of mass frac{10} | vedclass

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Question

(B) The moment of inertia $I$ of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2}M

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) The moment of inertia $I$ of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.Given $M = \frac{10}{\pi^2} \,kg$ and $R = 2 \,m$, we have $I = \frac{1}{2} 	imes \frac{10}{\pi^2} 	imes (2)^2 = \frac{20}{\pi^2} \,kg \cdot m^2$.The initial angular speed $\omega_1 = 90 \,rev/min = 90 	imes \frac{2\pi}{60} \,rad/s = 3\pi \,rad/s$.The final angular speed $\omega_2 = 120 \,rev/min = 120 	imes \frac{2\pi}{60} \,rad/s = 4\pi \,rad/s$.According to the work-energy theorem, the work done $W$ is equal to the change in rotational kinetic energy: $W = \Delta K = \frac{1}{2}I(\omega_2^2 - \omega_1^2)$.Substituting the values: $W = \frac{1}{2} 	imes \frac{20}{\pi^2} 	imes ((4\pi)^2 - (3\pi)^2) = \frac{10}{\pi^2} 	imes (16\pi^2 - 9\pi^2) = \frac{10}{\pi^2} 	imes 7\pi^2 = 70 \,J$.

$A$ thin uniform circular disc of mass $\frac{10}{\pi^2} \,kg$ and radius $2 \,m$ is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from $90 \,rev/min$ to $120 \,rev/min$ is (in $\,J$)

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