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A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I₁ is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I₂ is the moment of inertia with respect to an axis passing through one of the masses, it follows thata. l₁ = l₂. b. l₁ > l₂. c. l₂ > l₁.

Respuesta :

Answer:

option C

Explanation:

Mass of m is on the both the side of the dumbbell

moment of inertia of the object when the axis is passing through the center and perpendicular to it

distance from the center be r/2

I₁ = mr₁² + mr₂²

I₁ = [tex]m\dfrac{r^2}{4} + m\dfrac{r^2}{4}[/tex]

I₁ = [tex]\dfrac{mr^2}{2}[/tex]

when the axis pass through one mass

I₂ = mr₁² + mr₂²

r₁  = 0         r₂ = r

I₂ = m(0)² + m(r)²

I₂ = m r²

hence, I₂ > I₁

correct answer is option C

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