If the moment of inertia of an object about its center of mass is I_0, the momen

Question

If the moment of inertia of an object about its center of mass is I_0, the moment of inertia of the same object about an axis a distance l from the center of mass, is I" = I_0 + ml^2 where m is the mass of the object. Ex I_rot, 0 = 1/12 ML^2, I-rod, edge = 1/12 ML^2 + M(L/2)^2 = 1/3 ML^2 The moment of inertia of a sphere of mass M and radius R, a distance 1/2 R from the center of mass, is A 7/5 MR^2 B. 13/20 MR^2 C. 3/7 MR^2 D. MR^2 E. 1/2 MR^2 The moment of inertia of a lollypop of mass 10g and radius 2.0cm, with a stick of mass 6.0g and length 5.0cm, about the end of the stick, opposite to the candy, is given by A. 3.16 times 10^2kgm^2 B. 2.50 times 10^2kgm^2 C. 1.60 times 10^1kgm^1 D. 3.16 times 10^-5kgm^2 E. 1.60 times 10^-6kgm^2

Explanation / Answer

I) moment of inertia of solid sphere about an axis passing through center

IC = (2/5)MR^2

from parallel axis theorem

I = IC+ Md^2

d = R/2

I = (2/5)MR^2 +M(R/2)^2

I = (13/20)MR^2

correct option is (B)

(II)

m = 10 g , r = 2 cm , M = 6 g, L = 5 cm

moment of inertia of rod about one end = (1/12)mL^2

moment of inertia of lolly pop = (2/5)mr^2

I = (1/3)ML^2 +(2/5)mr^2 +mL^2

I= (1/3)(0.006*0.05^2) +(2*0.01*0.02^2/5) +(0.01*0.05^2)

I = 3.16*10^-5 kg.m^2

correct option is (D)

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