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2 years ago

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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 I_1 is the m

oment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and l_2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that l_1 = l_2. l_l > l_2. l_2 > l_1

1 answer:

Vanyuwa [196]2 years ago

7 0

Answer: $I_2>I_1$

Explanation:

Given

Shape of the object is dumbbell shaped

Moment of Inertia w.r.t an axis passing through center and perpendicular to it

$I_1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

$I_1=\frac{mr^2}{2}$

For the axis which passes through one of the masses

$I_2=mr^2$

so $I_2>I_1$

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