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 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 object. Ex. I_rod, 0 = 1/12 ML^3 I_road, = 1/12ML^2 + M(1/)^2 = 1/3 ML^3 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 10^2 kgm^2 B. 2.50 times 10^2 kgm^2 C. 1.60 times 10^1 kgm^2 D. 3.16 times 10^-5 kgm^2 E. 1.60 times 10^-6 kgm^2

Explanation / Answer

Part (i)-

Isphere,o  = 2/5 MR2  (About the center axis)

Then Momentum of inertia from distance R/2 = Isphere,o + M(R/2)2

= 2/5MR2 + 1/4MR2

= (2/5 + 1/4) MR2

=(13/20)MR2

Part (ii)-

Iend = [Irod,edge ] + [Isphere,o + M2(L+R)2]

= M1L2/3 + 2/5M2R2 + M2(0.07)2 (Where M1 = 0.006kg And M2 = 0.01kg, R = 0.02m, L = 0.05m)

=(0.006)(0.05)2 /3 + 2/5(0.01)(0.02)2 + (0.01)(0.07)2

= 5*10-6 + 1.6*10-6 + 49*10-6

= 55.6 *10-6 kgm2

There is no option related to this but this is the right answer.

Thank you !!! Have fun

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