Question

) 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 and perpendicular to the rod, it follows that

a. It depends on the values of m and r.

b. I1 > I2.

c. I2 > I1.

d. I1 = I2.

Answer 1

Formula for moment of inertia due to the point like object at the distance r from axis is mr²

Now for case 1

let rod is of length r

Now axis is passing through the centre of rod so distance of each dumbbell from axis is r/2

Now total moment of inertia = sum of moment of inertia due to both dumbbell.

So I​​​​​​1 = m(r/2)² + m(r/2)²

= mr²/2

Now for case 2

Axis is passing through one dumbbell as it's distance from axis is zero so it's contribution in moment of inertia is zero and other dumbbell is at distance r from axis

So total moment of inertia = moment of inertia of dumbbell at distance r.

I​​​​​​2 = mr​​​​​​​​​​²

So from this we can conclude that mr² >mr²/2

So I​​​​​​2 > I​​​​​​1.