A thin uniform bar has two small balls glued to its ends. The bar is 2.00m long

Question

A thin uniform bar has two small balls glued to its ends. The bar is 2.00m long and has a mass of 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses.

Find the moment of inertia of this combination about each of the following axes:

(a) an axis perpendicular to the bar through its center;

(b) an axis perpendicular to the bar through one of the balls.

For part b I've been trying to solve it using the parallel theorem but I'm getting wrong answer (checking from book)

I=Icm+Md^2 -> I=(ML^2/12)+(M+m1+m2)d^2->

I=(4*2^2/12)+[5.00(L/2)^2] -> I=16/12 +5 -> I= 6.33 kgm^2 though the correct answer is 7.33 where Am I wrong?

Explanation / Answer

I = moment of inertia of rod abt one end + moment of inertia of other mass

I = (Icm + Md2 )+ md2

I = (ML2/12) + Md2 + md2

I = 4 (2)2 /12 + (4) (1)2 + (0.5) (2)2

I = 7.33

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