Moment of inertia or **rotational inertia **(I) is equal to the product of **mass **(m) and the square of **distance **(r). Using the formula of rotational inertia: **I = m × r**** ^{2}**, the value of rotational inertia of an object can be calculated.

Let’s solve some problems based on this formula, so you’ll get a clear idea.

## Inertia Practice Problems

**Problem 1:** A disc having a mass of 4 kg is rotating about its center. If the distance from the axis of rotation is 5 m, then calculate the moment of inertia of a disc.

Solution:

Given data:

Mass of a disc, m = 4 kg

Distance from the axis of rotation, r = 5 m

Moment of inertia of a disc, I = ?

Using the formula of moment of inertia,

I = m × r^{2}

I = 4 × (5)^{2}

I = 4 × 25

I = 100 kg m^{2}

Therefore, the moment of inertia of a disc is **100 kg m ^{2}**.

**Problem 2:** Calculate the moment of inertia of a 250 gm ring rotating about its center. The distance from the axis of rotation is 6 m.

Solution:

Given data:

Moment of inertia of a ring, I = ?

Mass of a ring, m = 250 gm = 0.25 kg

Distance from the axis of rotation, r = 6 m

Using the formula of moment of inertia,

I = m × r^{2}

I = 0.25 × (6)^{2}

I = 0.25 × 36

I = 9 kg m^{2}

Therefore, the moment of inertia of a ring is **9 kg m ^{2}**.

**Problem 3:** Two balls A and B of mass 2 kg and 5 kg are connected by a rod of length 5 m and rotates about the axis CD. The distances of the two balls A and B from the axis of rotation are 2 m and 3 m. Calculate the moment of inertia of the system about the axis CD.

Solution:

Given data:

Mass of the ball A, m_{A} = 2 kg

Mass of the ball B, m_{B} = 5 kg

Distance of the ball A from the axis of rotation, r_{A} = 2 m

Distance of the ball B from the axis of rotation, r_{B} = 3 m

Moment of inertia of the system, I = ?

Using the formula of moment of inertia,

I = (m_{A} × r_{A}^{2}) + (m_{B} × r_{B}^{2})

I = (2 × 2^{2}) + (5 × 3^{2})

I = (2 × 4) + (5 × 9)

I = 8 + 45

I = 53 kg m^{2}

Therefore, the moment of inertia of the system is **53 kg m ^{2}**.

**Problem 4:** A ball of mass 300 gm is rotating on its own axis. One another ball of mass 400 gm is connected to it by the rod at a distance of 2 m. Calculate the moment of inertia of the system.

Solution:

Given data:

Mass of the ball 1, m_{1} = 300 gm = 0.3 kg

Mass of the ball 2, m_{2} = 400 gm = 0.4 kg

Distance of the ball 1 from the axis of rotation, r_{1} = 0 m

Distance of the ball 2 from the axis of rotation, r_{2} = 2 m

Moment of inertia of the system, I = ?

Using the formula of moment of inertia,

I = (m_{1} × r_{1}^{2}) + (m_{2} × r_{2}^{2})

I = (0.3 × 0^{2}) + (0.4 × 2^{2})

I = (0.3) + (0.4 × 4)

I = 0.3 + 1.6

I = 1.9 kg m^{2}

Therefore, the moment of inertia of the system is **1.9 kg m ^{2}**.

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