Q.A flywheel is a mechanical device specifically designed to efficiently store rotational energy. For a particular machine it is in the form of a uniform 20 kg disc of diameter 50 cm, able to rotate about its own axis. Calculate its kinetic energy when rotating at 1200 rpm. Use π² = 10. Calculate its moment of inertia, in case it is rotated about a tangent in its plane.

A uniform disc about its own axis has $I_z=\frac{1}{2}MR^2$, and the two theorems chain a diameter ($\frac{1}{4}MR^2$) to an in-plane tangent ($\frac{5}{4}MR^2$). …

A flywheel in the form of a uniform 20 kg disc of diameter 50 cm (R = 0.25 m) rotates about its own central axis at 1200 rpm (use $\pi^2=10$). Its rotational kinetic energy is found from $I_{own}=\frac{1}{2}MR^2$ and $\omega=2\pi n$, giving $KE=\frac{1}{2}I_{own}\omega^2=\frac{1}{2}\left(\frac{1}{2}MR^2\right)(2\pi n)^2=\pi^2MR^2n^2$, which evaluates to 5000 J. The second part finds the disc's moment of inertia about a TANGENT lying in its own plane: first the perpendicular-axes theorem gives the diametric moment of inertia $I_d=I_x=I_y=\frac{1}{2}I_{own}=\frac{1}{4}MR^2$ (since the own-axis and any two perpendicular diameters are mutually perpendicular and concurrent, and $I_x=I_y$ by symmetry); then, since a tangent in the …

For the tangent-in-plane value, go via radius of gyration: $K_d=R/2$ for a diameter, so $K_{tangent}^2=K_d^2+R^2=\frac{5}{4}R^2$ — …

Using the diameter 50 cm as R (R is 25 cm); applying parallel-axes directly from $I_z$ (the central transverse axis is NOT parallel to an in-plane tangent …