Rotational Motion

JEE Main level, numerical answer, medium

18 of 280

Question 18 of 280

Two identical thin rods of mass M kgM \mathrm{~kg} and length L mL \mathrm{~m} are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point PP and perpendicular to the plane of the rods is x12ML2 kg m2\frac{x}{12} \mathrm{ML}^{2} \mathrm{~kg} \mathrm{~m}^{2}. The value of xx is ____\_\_\_\_.

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Answer: 17

For vertical rod with axis at end P: I1=ML23I_1 = \frac{ML^2}{3}
For horizontal rod centered at distance L from P:
I2=∫−L/2L/2ML(x2+L2)dx=ML[x33+L2x]−L/2L/2I_2 = \int_{-L/2}^{L/2} \frac{M}{L}(x^2 + L^2)dx = \frac{M}{L}\left[\frac{x^3}{3} + L^2x\right]_{-L/2}^{L/2}
=ML[L312+L3]=13ML212= \frac{M}{L}\left[\frac{L^3}{12} + L^3\right] = \frac{13ML^2}{12}
Total: I=ML23+13ML212=4ML2+13ML212=17ML212I = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}
x=17x = 17

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