Lesson Video: Newto’s Second Law of Motion for Rotation in Terms of Moment of Inertia and Angular Acceleration | Nagwa Lesson Video: Newto’s Second Law of Motion for Rotation in Terms of Moment of Inertia and Angular Acceleration | Nagwa

# Lesson Video: Newto’s Second Law of Motion for Rotation in Terms of Moment of Inertia and Angular Acceleration Physics

In this video, we will learn how to calculate the torque on an object given its moment of inertia and its angular acceleration.

09:20

### Video Transcript

In this video, we will learn how to calculate the torque on an object given its moment of inertia and its angular acceleration. Before we apply Newton’s second law of motion to rotation, let’s first recall what Newton’s second law of motion is.

The equation that represents Newton’s second law of motion is that the net force, 𝐹 net, is equal to the mass of an object, 𝑚, times the acceleration of the object, 𝑎. When we study rotational motion, we use the rotational analog of this equation. The net torque on an object, 𝜏, is equal to the moment of inertia, 𝐼, times the angular acceleration, 𝛼.

Let’s compare our two equations to help us understand the rotational version a little bit better. The net force is the sum of all the individual forces acting on the object. In rotational motion, we use the net torque, which is the sum of all the individual torques acting on an object. Let’s refresh our memory on the common conventions to assign positive and negative values to certain directions. With linear forces, we usually assign to the right as positive and to the left as negative. With rotation, we usually assign a counterclockwise torque as positive and a clockwise torque as negative.

The rotational analog for mass is the moment of inertia, which is essentially the distribution of the mass throughout an object. When we look at the acceleration in rotational motion, we look at the angular acceleration. Now that we are familiar with Newton’s second law of motion for rotational motion, let’s add some values to our diagram to practice using the equation.

If we apply a net torque of six newton meters to an object that has a moment of inertia of three kilogram meters squared, what would be the angular acceleration of the object?

We begin by rearranging our formula to solve for 𝛼. To isolate 𝛼, we need to divide both sides by the moment of inertia, thereby canceling out 𝐼 on the right side of the equation. Our angular acceleration is equal to the net torque applied to the object divided by the moment of inertia of the object. We use negative six newton meters for the torque as we chose six newton meters as the magnitude of our torque and in our diagram we drew our torque to be clockwise implying that it must be negative.

For the moment of inertia, we chose a value of three kilogram meters squared. When we divide our two values, we get an angular acceleration of negative two rads per second squared. Let’s now take a look at this equation from a graphical standpoint. If we were to plot a graph of angular acceleration of an object against the torque applied to the object, what would that graph look like? It would look something like this, a diagonal line representing the direct linear relationship between torque applied to an object and the angular acceleration of the object.

What does the slope or gradient of this graph represent? The slope or gradient of a graph is equal to the change in 𝑦-value divided by the change in 𝑥-value. Looking back at how we labeled our axes, we can see that the 𝑦-axis is the angular acceleration 𝛼 and the 𝑥-axis is the torque or 𝜏. Returning to the formula at the top of our screen, we need to rearrange our variables, so we have an expression for 𝛼 divided by 𝜏. To do this, we multiply both sides of the equation by one over 𝐼 and one over 𝜏, canceling out the 𝜏 on the left side of the equation and canceling out the 𝐼 on the right side of the equation, leaving us with the expression 𝛼 over 𝜏 is equal to one over 𝐼, which means that the slope of our graph is one over the moment of inertia.

Let’s compare a couple different slopes and see what that tells us about our moment of inertia. We have added two more lines, a yellow line, which has a steeper slope than our original pink line, and a blue line, which has a more shallow slope than our original pink line. Which of the two new lines, yellow or blue, has a greater moment of inertia? To figure that out, let’s return back to the relationship we discovered between slope and moment of inertia. The moment of inertia is inversely proportional to the slope of our graph. This means that as the slope gets steeper, the moment of inertia gets smaller.

So, if the graph with the steepest slope has the smallest moment of inertia, then the graph with the shallowest slope will have the greatest moment of inertia. In this case, this would be our blue line, making our yellow line with the steepest slope have the smallest moment of inertia.

Let’s apply Newton’s second law of motion for rotation to a few example problems.

A spinning metal sphere has a moment of inertia of 1.7 kilograms meters squared. It has a constant angular acceleration of 2.0 radians per second squared. What is the magnitude of the torque on the object?

We have drawn a diagram of our spinning metal sphere with the information given from our problem, including angular acceleration and moment of inertia. To solve for the torque, we need to find a relationship between these three variables. We should recall Newton’s second law of motion for rotational motion, which is the net torque acting on an object, 𝜏 net, is equal to the moment of inertia of the object, 𝐼, times the angular acceleration of the object, 𝛼.

In our problem, we’re given the moment of inertia and we’re given the angular acceleration and asked to find the torque. Therefore, we do not need to rearrange our formula to solve for our unknown variable. Substituting in our values, we see that we are given 1.7 kilograms meter squared for our 𝐼 and we are given 2.0 radians per second squared for our 𝛼. Multiplying these two values together, we get a torque of 3.4 newtons times meters. The magnitude of the torque on the object is 3.4 newton meters.

The wheel of a train carriage has a moment of inertia of 28 kilogram meters squared. As the train is increasing in speed as it leaves the station, the angular acceleration of the wheel is 1.5 radians per second squared. What is the magnitude of the torque being applied to the wheel?

Drawing a diagram can help us visualize the situation. In our diagram, we chose one of the wheels of our train carriage and labeled it with the information from the problem. The moment of inertia, 𝐼, is given as 28 kilogram meters squared. The angular acceleration of the wheel is given as 1.5 radians per second squared. And we are trying to solve for the torque applied to the wheel.

We need an equation that relates these three variables together. We need to remember that Newton’s second law of motion applied to rotational motion is the net torque, 𝜏 net, is equal to the moment of inertia of the object, 𝐼, times the angular acceleration of the object, 𝛼. Looking at our problem, we are given 𝐼, 𝛼, and solving for 𝜏. Therefore, we do not need to rearrange our formula to solve for our unknown variable.

Substituting in our values, we have 28 kilogram meters squared for 𝐼 and 1.5 radians per second squared for 𝛼. When we multiply these two numbers together, we get a 𝜏 of 42 newton meters. The magnitude of the torque being applied to the wheel of the train carriage is 42 newton meters.

In a hard disk drive, a constant torque of 14.0 newton meters is applied to the magnetic disk when the drive starts recording data. The magnetic disk has a moment of inertia of 1.12 kilogram meters squared. What is the magnitude of the angular acceleration of the disk?

We can draw a diagram that represents the disk rotating with the information provided from the problem. In our diagram, we have labeled the magnitude of the torque applied to be 14.0 newton meters, the moment of inertia of the disk to be 1.12 kilogram meters squared, and we are solving for our angular acceleration 𝛼.

To solve the problem at hand, we needed an equation that relates these three variables together. Newton’s second law of motion when applied to rotational motion is the net torque, 𝜏 net, is equal to the moment of inertia of the object, 𝐼, times the angular acceleration of the object, 𝛼. The problem asks us to solve for the angular acceleration. However, our equation right now is written in the form that it’s solving for torque. Therefore, we must rearrange our formula to solve for 𝛼.

To isolate 𝛼, we must divide both sides of the equation by 𝐼, canceling out 𝐼 from the right side of the equation, leaving us with torque divided by moment of inertia is angular acceleration. Substituting in our values, we have 14.0 newton meters for our torque and 1.12 kilograms meters squared for our moment of inertia. When we divide these two numbers, we get an angular acceleration of 12.5 radians per second squared. The magnitude of the angular acceleration of the disk is 12.5 radians per second squared.

Key Points

Use 𝜏 equals 𝐼𝛼 to find the torque on an object given its moment of inertia and its angular acceleration. The slope of the graph of angular acceleration against torque is one over 𝐼.

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