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(B) The moment of inertia $I$ of the system about the $X$-axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-t

Vedclass · Accepted Answer

$184$

Vedclass · Answer

(B) The moment of inertia $I$ of the system about the $X$-axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the $X$-axis.Given masses and their positions on the $Y$-axis:$m_1 = 4.00 \, kg$ at $y_1 = 3.00 \, m$,so $r_1 = 3.00 \, m$.$m_2 = 2.00 \, kg$ at $y_2 = -2.00 \, m$,so $r_2 = 2.00 \, m$.$m_3 = 3.00 \, kg$ at $y_3 = -4.00 \, m$,so $r_3 = 4.00 \, m$.Calculating the moment of inertia:$I = (4.00 \, kg)(3.00 \, m)^2 + (2.00 \, kg)(2.00 \, m)^2 + (3.00 \, kg)(4.00 \, m)^2$$I = (4.00 	imes 9) + (2.00 	imes 4) + (3.00 	imes 16) = 36 + 8 + 48 = 92 \, kg \cdot m^2$.The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega^2$.Given $\omega = 2 \, rad/s$:$K.E. = \frac{1}{2} 	imes 92 	imes (2)^2 = \frac{1}{2} 	imes 92 	imes 4 = 184 \, J$.

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

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