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A composite rod comprising two rods of mass m and 2m and each of length l = 1 m as shown in the figure. Assume $ω$ = 10 rad/s and m = 1 kg, the KE of the rotating composite rod is

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250 J

125 J

175 J

150 J

answer is C.

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Detailed Solution

Moment of inertia of composite rod about P

$I_{P} = \frac{{ml}^{2}}{3} + \frac{(2 m) l^{2}}{12} + 2 m {(\frac{3}{2} l)}^{2} = 5 {ml}^{2}$

Total kinetic energy k total = $\frac{1}{2} I_{p} ω^{2}$

= $\frac{1}{2} (5 {ml}^{2}) ω^{2} = \frac{5}{2} {ml}^{2} ω^{2} = 250 J$

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