Moments of Inertia – formulas & sample numerical

Last updated on February 1st, 2022 at 05:40 pm

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 =mr²α . This is an important result because it relates to torque and angular acceleration. The quantity mr² is called the moment of inertia, I.

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 = mr²

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 = mr²
• Solid disk rotating around its center: I =(1/2) mr²
• Hollow cylinder rotating around its center (such as a car tire): I = mr²
• Solid cylinder rotating around its center: I = (1/2) mr²

• Hollow sphere rotating around any diameter: I = (2/3) mr²
• Solid sphere rotating around any diameter: I = (2/5) mr²
• Point mass rotating at radius r: I = mr²

• Rod of length L rotating around an axis perpendicular to it and through its center: I= (1/12) mL²
• Rod of length L rotating around an axis perpendicular to it and through one end: I = (1/3) mL²

• Rectangle rotating around an axis along one edge
  & L is the length of the other edge: I= (1/3) mL²

• Rectangle with sides r₁ and r₂ rotating around a perpendicular axis through the center,
  I = (1/12) m (r₁² + r₂²)

• 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) mL²

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/s², what torque is operating on it?

Solution:

m = 5kg

r = 10 cm = 0.1 m

α = 3 rad/s^2

Hence, I = (1/2) mr² = (1/2) .5. 0.1² = 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/s², what torque are you applying?

Solution:

m = 5 kg

r = 0.5 m

α = 4 rad/s^2

I = (2/5) mr² = (2/5) 5. 0.5² = 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/s², 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/s², 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/s². 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]