📂Classical Mechanics

Angular Momentum and Torque

Angular Momentum¹

Momentum is a physical quantity that represents the state of motion of a moving object. The larger the mass and the faster the speed, the greater the momentum. In physics, there is an interest in how the motion of an object changes. Therefore, the force, which is the cause of changing the state of motion of an object, is expressed as a change in momentum.

$$ \mathbf{F}=\frac{d \mathbf{p}}{dt} $$

Now, we are trying to define a similar physical quantity for rotational motion. In rotational motion, unlike translational motion, there is a radius of rotation which affects the motion of the body. Hence, the quantity that represents the state of motion of a rotating body is called angular momentum, and is defined as follows.

$$ \mathbf{L}=\mathbf{r}\times \mathbf{p} $$

Since linear momentum is the product of mass and velocity, it does not change with the origin, but angular momentum includes the position vector $\mathbf{r}$, so its value can change depending on where the origin is set. It should be noted that defining it this way is reasonable and natural upon verification of several facts.

Torque

In translational motion, the physical quantity representing the state of motion is momentum and the change in momentum is referred to as force. Similarly, in rotational motion, the physical quantity representing the state of motion is angular momentum, and the change in angular momentum can be something that changes the state of rotational motion. This physical quantity is called Torque and is denoted by $\mathbf{N}$.

$$ \mathbf{N}=\frac{ d \mathbf{L}}{ dt } $$

Expanding the right-hand side of the above equation yields the following.

$$ \begin{align*} \frac{ d \mathbf{L}}{ dt }&=\frac{ d (\mathbf{r}\times \mathbf{p})}{ dt } \\ &=\frac{d \mathbf{r}}{dt}\times \mathbf{p}+\mathbf{r}\times \frac{ d \mathbf{p}}{ dt } \\ &= \mathbf{v}\times \mathbf{p} + \mathbf{r}\times\mathbf{F} \end{align*} $$

Since $\mathbf{p} = m\mathbf{v}$, it follows that $\mathbf{v}\times \mathbf{p}=\mathbf{v}\times(m\mathbf{v})=\mathbf{0}$. Therefore, we obtain the equation below.

$$ \frac{ d \mathbf{L}}{ dt }=\mathbf{r} \times \mathbf{F} $$

According to the above equation, if the net force is $\mathbf{0}$, there is no change in angular momentum, meaning that the state of rotational motion does not change. It can be confirmed that the definition of angular momentum accurately reflects real physical phenomena. Furthermore, from the above equation, the formula for torque can be derived as follows.

$$ \mathbf{N}=\mathbf{r} \times \mathbf{F} $$

Consider a body moving in a straight line. Assume that an external force is applied in the same direction as the direction of movement. Then, since the direction is the same as those of $\mathbf{r}$ and $\mathbf{F}$, the torque is $\mathbf{0}$. Torque changes angular momentum, but if it is $\mathbf{0}$, it means that the state of the body’s rotational motion does not change. In reality, the state of a body moving in a straight line not rotating remains “not rotating.” Therefore, we can understand that this formula accurately represents real physical phenomena. Additionally, for bodies that rotate under a central force, since the direction of the position vector $\mathbf{r}$ and the central force $\mathbf{F}$ are the same, torque becomes $\mathbf{0}$, preserving angular momentum. This fact leads to the derivation of the law of areas. Torque is what causes changes in rotational motion, and how those changes are made is determined by the right-hand rule. As can be seen in the gif² above, if the torque direction is upwards, according to the right-hand rule, the status of the body changes to move counterclockwise.