Question

# Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:

A
15215MR2
B
1715MR2
C
13715MR2
D
20915MR2

Solution

## The correct option is A $$\dfrac{137}{15} MR^2$$For Ball using parallel axis theorem.$$I_{ball} = \dfrac{2}{5} MR^2 + M(2R)^2$$$$= \dfrac{22}{5} MR^2$$2 Balls so $$\dfrac{44}{5} MR^2$$Irod = for rod $$\dfrac{M(2R)^2}{R} = \dfrac{MR^2}{3}$$$$I_{system} = I_{Ball} + I_{rod}$$$$= \dfrac{44}{5} MR^2 + \dfrac{MR^2}{3}$$$$= \dfrac{137}{15} MR^2$$Physics

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