Did you know that every time we grocery shop, we experience Newton's second law? Before shopping, the empty cart effortlessly rounds the corners into the next aisle. However, after the cart becomes filled with groceries, it becomes hard to maneuver and you may find yourself struggling to make it to the next aisle. This is a direct result of Newton’s second law in angular form and the relationship between torque, moment of inertia, and angular acceleration. As you fill the cart, its moment of inertia increases, and as a result, you now must apply more torque to the cart for it to round the corners. Therefore, let us use this example as a starting point in understanding Newton's second law. In this article, we will define key concepts and work through some examples to aid in your understanding of Newton's second law.
Rotational Form of Newton's Second Law
Newton's second law in angular form corresponds to rotational motion and describes the relationship between torque, inertia, and angular acceleration. Its formula is the rotational analog of Newton's second law for linear motion, \( F=ma. \)
Rotational motion is a type of motion associated with objects traveling in a circular path.
Torque and Force
Torque is the rotational analog of force because it is the measure of the force required for an object to start rotating about an axis. As in linear motion where a force causes an object to accelerate, torque causes the angular acceleration of an object.
Torque can be defined by three formulas.
- Cross Product Formula,
- Magnitude Formula,
- Newtons Second Law Formula.
Cross-Product Formula
The cross-product definition of torque is expressed by the equation
$$\vec{\tau}=\vec{r} \times \vec{F}$$
where \( \vec{r} \) is the lever arm measured in \( \mathrm{m} \) and \( \vec{F} \) is the force applied measured in \( \mathrm{N}. \)
The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.
It is important to recognize that cross product is another term for vector product indicating that both \( r \) and \( F \) are vector quantities.
The direction of the vector resulting from the cross-product of two vectors is perpendicular to both vectors and therefore is normal to the plane defined by the two vectors.
Magnitude Formula
As a result of the cross-product, the magnitude definition of torque is expressed by the equation
$$\tau=rF\sin\theta$$
where \( r \) is the lever arm, \( F \) is the applied force, and \( \theta \) is the angle between the lever arm and the applied force. The variables, \( r \) and \( F \), no longer represent vectors but rather correspond to the magnitude of each vector.
Angular Acceleration and Linear Acceleration
Angular acceleration is the rotational equivalent of linear acceleration.
This relationship is expressed by the formula \( a=\alpha{r} \) which can be rewritten as \( \alpha=\frac{a}{r} \) where \( r \) represents radius.
Angular acceleration is an object's change in angular velocity with respect to time.
The mathematical formula corresponding to this definition is
$$\alpha=\frac{\Delta{\omega}}{\Delta{t}}$$ where \( \omega \) is angular velocity and \( t \) is time.
Moment of Inertia and Mass
The rotational analog of mass is inertia. An object's moment of inertia describes how mass is distributed relative to the axis of rotation. Meanwhile, mass itself describes the amount of matter within an object.
Formulas regarding an object's moment of inertia will vary depending on the shape of the object. For example, the moment of inertia of a point mass a distance \(r\) from the axis of rotation is given by \( I=mr^2. \) This formula is important because it is the basis for all other inertial formulas since objects can be constructed from sets of point masses.
Moment of inertia is the tendency for an object to resist a change in its rotational motion.
Formula of Newton's Second Law for Rotation
The equation for Newton's second law in angular form is expressed as
$$\tau =I\alpha$$
where \( \tau \) is the total net torque measured in \( \mathrm{N\,m} \), \( I \) is the moment of inertia measured in \( \mathrm{{kg}\,{m^2}} \) and \( \alpha \) is angular acceleration measured in \( \mathrm{\frac{rad}{s^2}}. \)
Meaning of Newton's Second Law in Angular Form
Newton's second law in angular form states that angular acceleration is proportional to torque and inversely proportional to the moment of inertia. This relationship can be seen if the equation is rewritten in terms of angular acceleration as \( \alpha= \frac{\tau}{I} \). This means that a greater torque implies a greater angular acceleration, and a greater moment of inertia implies a smaller angular acceleration. The latter is what happens with your shopping cart!
Proof of Newton's Second Law in Angular Form
To prove that Newton's second law can be written in angular form, let us consider this example. An object with mass \( m\) is rotating about an axis and a force is exerted at a distance \( r \) from the axis. We know that the object is constrained to rotate in a circular motion as a result of the fixed axis and we assume that the force is tangent to the circle. Therefore, using this information in conjunction with the relationship between linear and angular acceleration, \( a=\alpha{r} \), enables the equation to be rewritten as follows:
\[\begin{align*}F&=ma,\\ rF&=rma,\\ rF&=rm(\alpha r),\\ rF&=(mr^2)\alpha,\\ \tau&=I\alpha.\end{align*}\]
Thus, we can derive the angular form of Newton's second equation from the linear form.
Common Applications of Newton's Second Law
A common application of Newton’s second law in angular form is a merry-go-round.
Figure 2: A merry-go-round as an example of Newton's second law in angular form.
If you have ever taken a child to the park, you have demonstrated Newton's second law in angular form by using the merry-go-round. Rotating an empty merry-go-round is much easier than rotating one with children on it. Why? Well, the answer is Newton's second law in angular form. When we spin the merry-go-round, we grab its edge and begin to apply a perpendicular force to the radius of the merry-go-round. This allows us to apply the maximum amount of torque necessary for the merry-go-round to accelerate. However, as soon as more mass is added, i.e., children climb onto it, the merry-go-round accelerates less easily because the additional mass results in a larger moment of inertia. If we wanted to increase the angular acceleration, we would have to apply more torque (and thus more force if we keep grabbing the edge).
Torque Examples with Newton's Second Law in Angular Form
To solve problems regarding Newton's second law in angular form, one can use the equation for torque and apply it to different problems. As we have defined torque and discussed its relation to rotational motion as well as multiple variables, let us work through some examples to gain familiarity with the equations.
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 are needed.
- Apply the necessary formulas and solve the problem.
- Draw a picture if necessary to provide a visual aid.
Examples
Let us apply our newfound knowledge of Newton's second law in angular form to some examples.
An object, whose moment of inertia is \( 47.00\,\mathrm{{kg}\,{m^2}} \), rotates with an angular acceleration of \( 5.800\,\mathrm{\frac{rad}{s^2}}. \) Calculate the total net torque that the object is experiencing.
After reading the problem, we write down what we are given:
- angular acceleration,
- moment of inertia.
Therefore, applying the equation for torque expressed in the form of Newton's second law in angular form, our calculations will be as follows:
$$\begin{align}\tau &= I\alpha\\\tau &= \left(47.00\,\mathrm{{kg}\,{m^2}}\right)\left(5.800\,\mathrm{\frac{rad}{s^2}}\right)\\\tau &= 272.6\,\mathrm{N\,m}\\\end{align}$$
The amount of torque the object is experiencing is \( 272.6\,\mathrm{N\,m}.\)
Now, let's complete a similar example where we solve for the moment of inertia rather than torque.
If \( 2.1\,\mathrm{N\,m} \) of torque is applied to a point mass whose angular acceleration is \( 0.87\,\mathrm{\frac{rad}{s^2}}. \) Calculate the moment of inertia of the point mass. If the mass of the point mass is given to be \( 1.1\,\mathrm{kg}, \) calculate the distance of the point mass to the axis of rotation.
After reading the problem, we write down what we are given:
- torque,
- angular acceleration,
- mass.
Therefore, applying the equation for torque expressed in the form of Newton's second law in angular form, our calculations will be as follows:
$$\begin{align}\tau &= I\alpha\\\frac{\tau}{\alpha} &=I\\\frac{2.1\,\mathrm{N\,m}}{0.87\,\mathrm{\frac{rad}{s^2}}} & = I\\ I & =2.4\,\mathrm{kg\,m^2}.\\\end{align}$$
Now insert this value into the equation corresponding to the moment of inertia for a point mass, and we can calculate the radius of the point mass as follows:
$$\begin{align}I &=mr^2\\2.4\,\mathrm{kg\,m^2}&=(1.1\,\mathrm{kg})r^2\\\frac{2.4\,\mathrm{kg\,m^2}}{1.1\,\mathrm{kg}} &= r^2\\ r^2 &=2.2\,\mathrm{m^2}\\r&=\sqrt{2.2\,\mathrm{m^2}}=1.5\,\mathrm{m}.\\\end{align}$$
We conclude that the distance of the point mass to the axis of rotation is \( 1.5\,\mathrm{m}. \)
Finally, let's end with a slightly more complicated example using the moment of inertia formula corresponding to a solid sphere.
A bowling ball, with a mass of \( 1.7\,\mathrm{kg} \) and a radius of \( 0.76\,\mathrm{m} \), rotates with an angular acceleration of \( 3.3\,\mathrm{\frac{rad}{s^2}}. \) Calculate the net torque exerted on the bowling ball. Note that a bowling ball is considered a solid sphere so its moment of inertia is given by \( I=\frac{2}{5}m{r^2}. \)
After reading the problem, we write down what we are given:
- mass,
- radius,
- angular acceleration.
Before solving for torque, we must first calculate the moment of inertia for the bowling ball. Using the formula for the moment of inertia of a solid sphere and the corresponding values, our calculations are:
$$\begin{align}I &=\frac{2}{5}m{r^2}\\I &=\mathrm{\frac{2}{5}\left(1.7\,\mathrm{kg}\right)\left({0.76\,\mathrm{m}}\right)^2}\\I&=0.39\,\mathrm{{kg}\,{m^2}}.\\\end{align}$$
Now solving for torque, our calculations are:
$$\begin{align}\tau &= I\alpha\\\tau &= \left(0.39\,\mathrm{{kg}\,{m^2}}\right)\left(3.3\,\mathrm{\frac{rad}{s^2}}\right)\\\tau &= 1.3\,\mathrm{N\,m}.\\\end{align}$$
We conclude that the amount of torque needed to give the object the given angular acceleration about the specified axis is \( 1.3\,\mathrm{N\,m}. \)
Newton's Second Law in Angular Form - Key takeaways
- Newton's second law is represented by the formula \( F=ma. \)
- Newton's second law in angular form describes the relationship between torque, inertia, and angular acceleration.
- Newton's second law in angular form is represented by the formula \( \tau=I{\alpha}. \)
- The formula for Newton's second law in angular form can be derived from Newton's second law for linear motion.
- Newton's second law can be written in angular form because torque, inertia, and angular acceleration are equivalents to force, mass, and acceleration.
- A greater torque implies a greater angular acceleration, and a greater moment of inertia implies a smaller angular acceleration.
- A common application of Newton’s second law in angular form is a merry-go-round.
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