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(D) The rotational kinetic energy is given by $K = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the axis of rotation.$1$. For rotati

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All of the above.

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(D) The rotational kinetic energy is given by $K = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the axis of rotation.$1$. For rotation about the $y$-axis: The masses $m$ are on the $y$-axis,so their distance from the axis is $0$. The masses $M$ are at distance $a$ from the $y$-axis. Thus,$I_y = M(a)^2 + M(a)^2 = 2Ma^2$. The kinetic energy is $K_y = \frac{1}{2}(2Ma^2)\omega^2 = Ma^2\omega^2$. This is independent of $m$.$2$. For rotation about the $z$-axis: The masses $M$ are at distance $a$ from the $z$-axis,and masses $m$ are at distance $b$ from the $z$-axis. Thus,$I_z = M(a)^2 + M(a)^2 + m(b)^2 + m(b)^2 = 2Ma^2 + 2mb^2$. The kinetic energy is $K_z = \frac{1}{2}(2Ma^2 + 2mb^2)\omega^2 = (Ma^2 + mb^2)\omega^2$. This depends on $m$.$3$. Since the moment of inertia $I$ depends on the axis of rotation,the rotational kinetic energy $K = \frac{1}{2}I\omega^2$ also depends on the axis of rotation.Therefore,all statements are correct.

Four point masses are fastened to the corners of a frame of negligible mass lying in the $xy$ plane as shown in the figure. Let $\omega$ be the angular speed of rotation. Then:

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