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, it follows that

Answer 1

Options:

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

b. I1 > I2.

c. I2 > I1.

d. I1 = I2.

Answer:

c) I​​​​​​₂ > I​​​​​​₁.

Step-by-step explanation:

The moment of inertia is given by I = mr²

When the axis is passing through the center of the rod:

length of the rod = r

Distance of each of the dumbbell from the axis = r/2

Since both dumbbells are of the same mass, m

Moment of inertia due to dumbbell 1 = m(r/2)²

Moment of inertia due to dumbbell 2 = m(r/2)²

Total moment of inertia = Moment of inertia due to dumbbell 1 + Moment of inertia due to dumbbell 2

I​​​​​​₁ = m(r/2)² + m(r/2)²

I​​​​​​₁ = mr²/2 ....................(1)

Now for case 2

When the axis passes through one dumbbell

Distance of the dumbell from axis, r = 0, therefore moment of inretia = 0

Distance of the second dumbbell from the axis = r

Total moment of inertia = moment of inertia of the second dumbbell at distance r.

I​​​​​​₂ = mr​​​​​​​​​​² .........................(2)

From equations, (1) and (2), it is obvious that mr² >mr²/2

i.e. I​​​​​​₂ > I​​​​​​₁.

Answer 2

I₁ is less than I₂

Step-by-step explanation:

Here is the complete 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 is I1 greater than, equal to, less than, or noncomparable to I2?

Solution

The moment of inertia I₁ about an axis through the center of the rod = m₁r₁² + m₂r₂² . Since m₁ = m₂ = m and r₁ = r₂ = r/2 where r₁ and r₂ are the distances of m₁ and m₂ from the axis passing through the center of the rod and r is the length of the rod.

I₁ = m₁r₁² + m₂r₂² = I₁ = m(r/2)² + m(r/2)² = mr²/2

The moment of inertia I₂ about an axis passing through one of the masses = m₁r₁² + m₂r₂² . Since m₁ = m₂ = m and r₁ = 0 and r₂ = r where r₁ and r₂ are the distances of m₁ and m₂ from the axis passing through m₁ and r is the length of the rod.

I₂ = m₁r₁² + m₂r₂² = m(0)² + m(r)² = mr²

Since I₁ = mr²/2 and I₂ = mr², it follows that I₁ = I₂/2.

So I₁ is less than I₂