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

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

A I total = Irod + 2*Iball = ML^2 /12 + 2* m(L/2)^2 = 9*2^2 /12 + 2*0.3*1^2 = 3.6 kg*m^2 B) I total = Irod + 1*Iball = ML^2 /3 + m(L)^2 = 9*2^2 /3+ 0.3 *2^2 = 13.2 kg.m^3 C) 0 D) Itotal = 9*0.5^2 + 2* 0.3 *0.5^2 = 2.4 kg.m^2

ID: 1445423 • Letter: A

Question

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 9.00 kg , while the balls each have mass 0.300 kg and can be treated as point masses. A) Find the moment of inertia of this combination about an axis perpendicular to the bar through its center. B) Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls. Express your answer with the appropriate units. C) Find the moment of inertia of this combination about an axis parallel to the bar through both balls. D) Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

Explanation / Answer

I total = Irod + 2*Iball = ML^2 /12 + 2* m(L/2)^2

= 9*2^2 /12 + 2*0.3*1^2 = 3.6 kg*m^2

B) I total = Irod + 1*Iball = ML^2 /3 + m(L)^2

= 9*2^2 /3+ 0.3 *2^2 = 13.2 kg.m^3

C) 0

D) Itotal = 9*0.5^2 + 2* 0.3 *0.5^2 = 2.4 kg.m^2

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

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

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