Rotational Motion

JEE Main level, numerical answer, medium

9 of 280

Question 9 of 280

Moment of inertia about an axis ABAB for a rod of mass 4040 kg and length 33 m is same as that of a solid sphere of mass of 1010 kg and radius RR about an axis parallel to ABAB axis with separation of 33 m as shown in figure below. The value of RR is given as α2\sqrt{\dfrac{\alpha}{2}}. The value of α\alpha is _______.

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Answer: 15

Moment of inertia of the rod about axis ABAB passing through one of its ends:
Irod=ML23=40×(3)23=40×93=120 kg⋅m2I_{\text{rod}} = \dfrac{ML^2}{3} = \dfrac{40 \times (3)^2}{3} = \dfrac{40 \times 9}{3} = 120\text{ kg}\cdot\text{m}^2

Moment of inertia of the solid sphere about the same axis ABAB, using the parallel axis theorem I=Icm+md2I = I_{\text{cm}} + md^2:
Mass m=10 kgm = 10\text{ kg}, radius RR, distance from the centre of the sphere to axis ABAB is d=3 md = 3\text{ m}.

Isphere=25mR2+md2=25(10)R2+10(3)2=4R2+90I_{\text{sphere}} = \dfrac{2}{5}mR^2 + md^2 = \dfrac{2}{5}(10)R^2 + 10(3)^2 = 4R^2 + 90

Equating the two moments of inertia:
120=4R2+90120 = 4R^2 + 90
4R2=304R^2 = 30
R2=152R^2 = \dfrac{15}{2}
R=152R = \sqrt{\dfrac{15}{2}}

Comparing with R=α2R = \sqrt{\dfrac{\alpha}{2}}:
α=15\alpha = 15

Hence, α=15\alpha = 15.

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