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Torque and Rotational Motion

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Torque is a word you may have heard but don’t fully understand. Well, what if I told you, even without understanding it, an average person demonstrates the concept of torque over 7,000 times a year by traveling to and from home, work, and any other activities throughout their day by the simple act of opening a door. This sounds crazy, right? Well, let’s take a dive deeper into this and learn how that’s possible.

Example of Torque: Door Hinge, adapted from public domain image.

This article will introduce the concepts of torque and rotational motion. Before going through some examples, we will define torque and rotational motion and how they are related to one another and angular momentum.

Below we will look at a couple of definitions and descriptions applicable to torque and rotational motion.

The definition of torque is as follows:

The symbol \(\tau\) represents **torque**, which is the amount of force applied to an object that will cause it to rotate about an axis.

The mathematical formula for torque is:

$$ \tau = r F \sin{\theta} $$

where \(r\) represents the radius measured in meters, \(\mathrm{m}\), \(F\) represents force measured in newtons, \(\mathrm{N}\), and \(\theta\) represents angular displacement. The SI unit for torque is newton meter, \(\mathrm{N}\,\mathrm{m}\).

As torque is a vector, it has magnitude and direction, where its direction can either be clockwise or counterclockwise. The amount of torque applied to an object will always depend on two factors:

- How much force is applied
- The perpendicular distance from the axis of rotation

In rotational motion, the components of velocity, acceleration, and displacement are the same form as their linear motion equivalents; however, we define them in terms of variables associated with rotational motion.

**Rotational motion **is a type of motion associated with objects traveling in a circular path. The force which causes these objects to travel in a circular path is known as torque.

We show the formulas for angular velocity, angular acceleration, and angular displacement below:

$$ \omega=\frac{\theta}{t} $$

where angular velocity is measured in radians per second, \(\frac{\mathrm{rad}}{\mathrm{s}}\).

The derivative yields the equation $$\omega=\frac{d\theta}{dt}$$, which is the definition of instantaneous velocity.

$$ \alpha=\frac{\omega}{t} $$

where angular acceleration is measured in radians per second squared, \(\frac{\mathrm{rad}}{\mathrm{s}^2}\).

The derivative yields the equation $$\alpha=\frac{d\omega}{dt}$$, which is the definition of instantaneous acceleration.

$$ \theta = \omega t $$

where we measure angular displacement in radians, \(\mathrm{rad}\).

In linear motion, we know that objects move due to force. However, the force causing objects to move in rotational motion is known as torque. As a result, we can write the equation for torque in the same form as Newton’s second law, \(F=ma\), and we show the formula for torque below:

$$ \tau = I\alpha $$

where \(I\) is the moment of inertia and \(\alpha\) is angular acceleration.

Besides solving problems with the equation for rotational torque, we can also use it to determine the relationship between variables simply by rearranging terms. For example, if asked what the relationship between angular acceleration and torque is, one can rearrange this equation to solve for angular acceleration and get the following:

$$ \alpha = \frac{\tau}{I} $$

As a result, we can determine that angular acceleration is proportional to torque and inversely proportional to the moment of inertia. Now, we can go one step further by rewriting variables in terms of other variables. For example, if we know angular acceleration is equal to \(\alpha = \frac{\omega}{t}\) and insert this into the equation for torque, we will get the following:

$$ \tau = I \frac{\omega}{t} = \frac{I \omega}{t} $$

As a result, we can see another recognizable term associated with rotational motion. The term \(I\omega\) represents angular momentum. Therefore, we can rewrite the equation for torque in terms of angular momentum as follows:

$$ \tau = \frac{L}{t} $$

Therefore, without doing any mathematical calculations, we can determine the different relationships between variables associated with rotational motion.

People demonstrate the concept of torque almost every day of their lives and may not even know it. Every time we open a door, we use the concept of torque as we cause the door to rotate on its hinges. From the formula for torque, as defined above, we know that torque is directly related to radius and force. Using this knowledge, we can understand why we place door handles at the farthest point from the door hinges. Let’s say that it requires \( 100\,\mathrm{N}\,\mathrm{m}\) of torque to open a door, and the distance from the hinges to the handle is \( 2\,\mathrm{m}\). We can then conclude that it would require \(50\,\mathrm{N}\) of force to open the door. Now, if we move the door handle to the center of the door, the radius becomes \(1\,\mathrm{m}\), and we would then have to apply a force of \(100\,\mathrm{N}\) to open the door. This change demonstrates why radius is important to torque and why door handles are located at the farthest point from the hinges. Door handles at the farthest point allow for a maximum radius, allowing us to open doors with ease as we can apply less force. Door handles located in the center of a door would make opening doors harder for us because a smaller radius means we have to use more force for the door to open.

To solve torque and rotational motion problems, the equation for torque can applied to different problems. As we have defined torque and discussed its relation to rotational motion, let us work through some examples to gain a better understanding of total mechanical energy. Note that before solving a problem, we must always remember these simple steps:

- Read the problem and identify all variables given within the problem.
- Determine what the problem is asking and what formulas apply.
- Apply the necessary formulas to solve the problem.
- Draw a picture if necessary to provide a visual aid

Using these steps, now let us work through some examples.

A plumber uses a \(2.5\,\mathrm{m}\) wrench to loosen a bolt. If he applies \(65\,\mathrm{N}\) of force, calculate how much torque is needed to loosen the bolt.

After reading the problem, we are asked to calculate the torque needed to loosen a bolt and are given the radius of the wrench as well as the amount of force being applied. Therefore, using the formula for torque, our calculations are as follows:

$$ \begin{aligned} \tau &= rF\sin{\theta} \\ \tau &= \left(2\,\mathrm{m}\right)\left(65\,\mathrm{N}\right)\sin{90} \\ \tau &= 130\,\mathrm{N}\,\mathrm{m} \end{aligned} $$

The amount of torque needed to cause the wrench to rotate and loosen the bolt is \(130\,\mathrm{N}\,\mathrm{m}\).

Note that the plumber is applying a perpendicular force to the wrench, thus creating a \(90^{\circ}\) angle.An object, whose moment of inertia is \(45\,\mathrm{kg}/\mathrm{m}^2\), rotates with an angular acceleration of \(3\,\mathrm{rad}/\mathrm{s}^2\). Calculate the torque needed for this object to rotate about an axis.

After reading the problem, we are asked to calculate the torque needed for an object to rotate about an axis and are given the object's angular acceleration and moment of inertia. Therefore, using the formula for torque, our calculations are as follows:

$$ \begin{aligned} \tau &= I\alpha \\ \tau &= \left( 45\,\mathrm{kg}/\mathrm{m}^2 \right) \left( 3\,\mathrm{rad}/\mathrm{s}^2 \right) \\ \tau &= 135\,\mathrm{N}\,\mathrm{m} \end{aligned} $$

The amount of torque needed to rotate the object about an axis is \(135\,\mathrm{N}\,\mathrm{m}\).

- Torque is the force needed for an object to rotate about an axis.
- Rotational motion is the motion of objects traveling in a circular path.
- Rotational motion is associated with angular velocity, \(\omega\), angular acceleration, \(\alpha\), and angular displacement, \(\theta\).
- We write the formula for torque in terms of radius and force, \(\tau = r F \sin{\theta}\), and angular acceleration and moment of inertia, \(\tau = I\alpha\).
- Torque occurs in our everyday life when we open doors.

The unit of torque in rotational motion is newton meter.

torque equals force times distance

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