Three particles are situated on a light and r | vedclass

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Question

Mass of first object, $m_{1}=4.00 \mathrm{kg}$

Mass of second object, $m_{2}=2.00 \mathrm{kg}$

Mass of third object, $m_{3}=3.00 \mathrm{kg}$

Vedclass · Accepted Answer

$184$

Vedclass · Answer

Mass of first object, $m_{1}=4.00 \mathrm{kg}$

Mass of second object, $m_{2}=2.00 \mathrm{kg}$

Mass of third object, $m_{3}=3.00 \mathrm{kg}$

Distance of first object from $x$ $-axis$, $r_{1}=3.00 \mathrm{m}$

Distance of second object from $x$ $-$axis, $r_{2}=-2.00 \mathrm{m}$

Distance of third object from $\mathrm{x}$ $-axis,$ $r_{3}=-4.00 \mathrm{m}$

Angular velocity, $\omega=2 \mathrm{rad} / \mathrm{s}$

$I=\sum_{1}^{n} m_{i} r_{i}^{2}$

$I=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}$

$I=(4.00 \mathrm{kg})(3.00 \mathrm{m})^{2}+(2.00 \mathrm{kg})(-2.00 \mathrm{m})^{2}+(3.00 \mathrm{kg})(-4.00 \mathrm{m})^{2}$

$I=92 \mathrm{kg} . \mathrm{m}^{2}$

$K \cdot E_{	ext {rotational }}=\frac{1}{2} I \omega^{2}$

$K . E_{	ext {rotational }}=\frac{1}{2}\left(92 \mathrm{kg} . \mathrm{m}^{2}ight)(2 \mathrm{rad} / \mathrm{s})^{2}=184 \mathrm{J}$

Three particles are situated on a light and rigid rod along $Y$axis as shown in the figure. If the system is rotating with an angular velocity of $2\,rad/\sec $about $X$axis, then the total kinetic energy of the system is ...... $J$

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