The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc $=5 \mathrm{~kg}$, radius of disc $=1 \mathrm{~m}$ ) approximately

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Aicharya · Accepted Answer

$\mathrm{KE}=\frac{1}{2} I \omega^2=\frac{1}{2} \times \frac{3}{2} m r^2 4 \pi^2 \mathrm{f}^2$

$\begin{aligned} & \Rightarrow \quad \mathrm{KE}=3 m r^2 \times \pi^2 \mathrm{f}^2=3 \times 5 \times 1 \times 10 \times 1 \\ & \mathrm{KE}=150 \mathrm{~J}\end{aligned}$

The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a…

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