In this post, we will focus on the formulas of the **moment of inertia** and also will solve a few interesting sample numerical problems using the moment of inertia formulas.

## Moments of Inertia – concepts & definition

The torque equation gives us: **τ =mr ^{2}α **. This is an important result because it relates to torque and angular acceleration. The quantity

**mr**is called the moment of inertia, I.

^{2}*And, the moment of inertia represents the effort we need to get something to change its angular velocity. *

So this equation of torque is usually written as **τ =Iα**

This equation is a general result, but the moment of inertia (**I**) differs depending on the situation. For example, * I* is different when you’re spinning a solid cylinder versus a solid sphere.

For a single small mass, **I = mr ^{2}**

By treating each mass as a collection of small masses, the moments of inertia for several other shapes have been figured out; some of them are listed in the section below.

## Moments of Inertia – formulas

In the following formulas, m is the total mass of the object, and r is always the radius (of the disk, cylinder, sphere, or hoop). L is the length of the rod or the length of the rectangle in the direction perpendicular to the axis of rotation.

- Hoop rotating around its center (like a bicycle tire):
**I = mr**^{2} - Solid disk rotating around its center:
**I =(1/2) mr**^{2} - Hollow cylinder rotating around its center (such as a car tire):
**I = mr**^{2} - Solid cylinder rotating around its center:
**I =****(1/2)**mr^{2}

- Hollow sphere rotating around any diameter:
**I = (2/3) mr**^{2} - Solid sphere rotating around any diameter:
**I = (2/5) mr**^{2} - Point mass rotating at radius r:
**I = mr**^{2}

- Rod of length L rotating around an axis perpendicular to it and through its center: I=
**(1/12) mL**^{2} - Rod of length L rotating around an axis perpendicular to it and through one end: I =
**(1/3) mL**^{2}

- Rectangle rotating around an axis along one edge

& L is the length of the other edge: I=**(1/3) mL**^{2}

- Rectangle with sides r
_{1}and r_{2}rotating around a perpendicular axis through the center,**I = (1/12) m (r**_{1}^{2}+ r_{2}^{2})

- Rectangle rotating around an axis parallel to one edge (L being the length of the other edge)

and the axis is passing through the center : I =**(1/12) mL**^{2}

## Moments of Inertia – sample numerical problems

Q 1)

A solid cylinder with a mass of 5.0 kg is rolling down a ramp.

If it has a radius of 10 cm and an angular acceleration of 3.0 radians/s^{2}, what torque is operating on it?

**Solution**:

m = 5kg

r = 10 cm = 0.1 m

**α** = 3 rad/s^2

Hence, I = (1/2) mr^{2} = (1/2) .5. 0.1^{2} = 0.025 kgm^2

Torque = I **α** = 0.025 x 3 Nm= 0.075 Nm

Q 2)

You’re spinning a 5.0-kg solid ball with a radius of 0.5 m. If it’s accelerating at 4.0 radians/s^{2}, what torque are you applying?

Solution:

m = 5 kg

r = 0.5 m

**α** = 4 rad/s^2

I = (2/5) mr^{2} = (2/5) 5. 0.5^{2} = 0.5 Kgm^2

Torque = T = I **α** = 0.5 x 4 Nm= 2 Nm

3)

A tire with a radius of 0.50 m and a mass of 1.0 kg is rolling down a street. If it’s accelerating with an angular acceleration of 10.0 radians/s^{2}, what torque is operating on it?

4) You’re spinning a hollow sphere with a mass of 10.0 kg and a radius of 1.0 m. If it has an angular acceleration of 15 radians/s^{2}, what torque are you applying?

5) You’re throwing a 300.0-g flying disc with a radius of 10 cm, accelerating it with an angular acceleration of 20.0 radians/s^{2}. What torque are you applying?

6) If you’re spinning a 2.0-kg solid ball with a radius of 0.5 m, starting from rest and applying a 6.0 N-m torque, what is its angular speed after 60.0 seconds?

7) If you’re spinning a 2.0-kg hollow ball with a radius of 0.50 m, starting from rest and applying a 12.0 N-m torque, what is its angular speed after 10.0 seconds?

[ Solution coming soon]