Some notes on Rotational Dynamics

Rotational Dynamics

Torque

In this section, we will see interaction of bodies with other objects while rotating and effects of those interactions. Let’s start with the simplest case: turning. While studying linear motion, we have seen that the cause of motion is force and it is a result of interaction between objects. The rotational equivalent of force is called torque and it depends on three things, one - the amount of force used, two - the distance from the axis of rotation, and three - the inclination of the force on the object(the angle between the force and the axis). \begin{equation}tau=rFsin\theta\end{equation} We can also use vector product to define force. It is the vector product between r and F(vector product, hence the order is important). It is also important to notice why the equation below is in boldface while the above is not. Recall that we represent vectors using boldface. \(\boldsymbol{\tau} = \textbf{r}\times\textbf{F}\)

Rotation as Seen by Newton’s Second Law

In linear motion, Newton’s second law states that if an object is accelerating, there is a net force on it or vice versa. It is a bi-implication between force and acceleration. \(\textbf{F}\iff \textbf{a}\) And we have the following as well: \(\textbf{F}=m\textbf{a}\) We have seen that Torque is the rotational equivalent of Force and angular acceleration is the rotational equivalent of acceleration(linear). What is then, the rotational equivalent of mass? \(\boldsymbol{\tau}=?\times\boldsymbol{\alpha}\) This unknown physical quantity is called the moment of inertia(I)

Moment of Inertia

In the simplest case possible, we have the following be true: \(\boldsymbol{\tau}=rF\) \(\boldsymbol{\tau}=r(m\textbf{a})\) \(\boldsymbol{\tau}=r(mr\boldsymbol{\alpha})\) \(\boldsymbol{\tau}=mr^2\boldsymbol{\alpha}\) Thus, our unknown in the above section is $\boldsymbol{mr^2}$. This physical quantity is called the rotational inertia of a point mass rotating about an axis r meters far from it and with a mass m. We use the symbol I to denote moment of inertia. Thus, for a point mass m rotating about an axis at a distance r, we have the moment of inertia be: \(I=mr^2\) Thus, our torque equation becomes: \(\tau=I\alpha\) Moment of inertia for a point mass and a mass-system with a simple structure is easy to compute. We just add the individual moments to get the total moment of inertia of the system. \(I=\sum_{1}^{n}{m_ir_i^2}\) However, for objects such as a ball and a rod, we can use simple calculus to compute their moments of inertia. Let’s start with a rod of mass M and length L that has a uniform linear mass density lambda. It is imperative to know which axis we are using to rotate the rod to compute its moment of inertia. Case 1 - rotation about its midpoint

\(\lambda=\dfrac{M}{L}=\dfrac{dm}{dr}\) We have seen that \(I=\sum_{1}^{n}{m_ir_i^2}\) That means, it is integration(since we sum up individual elements) \(I=\int_{\frac{-L}{2}}^{\frac{L}{2}}{r^2dm}\) \(\lambda=\frac{dm}{dr}\iff dm=\lambda dr\) \(I=\int_{\frac{-L}{2}}^{\frac{L}{2}}{r^2(\lambda dr)}\) \(I=\lambda\dfrac{r^3}{3}\Biggr|_\frac{-L}{2}^{\frac{L}{2}}\) \(I=\lambda\dfrac{(\frac{L}{2})^3}{3}-\lambda\dfrac{(\frac{-L}{2})^3}{3}\) \(I=\lambda\dfrac{L^3}{12}\) Since $\lambda=\dfrac{M}{L}$, we then get: \(I=\dfrac{ML^2}{12}\) Case 2 - rotation about one end \(I=\int_{0}^{L}{r^2dm}\) \(\lambda=\frac{dm}{dr}\iff dm=\lambda dr\) \(I=\int_{0}^{L}{r^2(\lambda dr)}\) \(I=\lambda\dfrac{r^3}{3}\Biggr|_{0}^{L}\) \(I=\lambda\dfrac{L^3}{3}-\lambda\dfrac{0^3}{3}\) \(I=\lambda\dfrac{L^3}{3}\) Since $\lambda=\dfrac{M}{L}$, we then get: \(I=\dfrac{ML^2}{3}\) We see that the axis of rotation of an object actually matters and affects the moment of inertia of an object.

Work Done and Rotational Kinetic Energy

We have seen in linear cases how to calculate work done and kinetic energy. While studying work, we had: \(W=\textbf{F}\cdot\textbf{r}\) We can calculate work by substituting the rotational elements of F and r which are $\tau$ and $\theta$ respectively. Thus, \(W=\boldsymbol{\tau}\cdot\boldsymbol{\theta}\) Similarly for kinetic energy, we have seen that: \(\text{KE}=\dfrac{1}{2}mv^2\) When we substitute the rotational equivalents of m and v into the equation, we get the following: \(KE_{rot}=\dfrac{1}{2}I\omega^2\) An object rolling down an inclined plane, for example, has both a rotational kinetic energy and rotational kinetic energy. The existence of one doesn’t imply or depend on the existence of the other.

Work-KE Theorem for Rotational Motion?

We have seen above that: \(\text{W}=\tau\theta\) \(\text{W}=I\alpha\theta\) \(\text{W}=I(\dfrac{w_f^2-w_i^2}{2})\) \(\text{W}=\dfrac{Iw_f^2}{2}-\dfrac{Iw_i^2}{2}\) \(\text{W}=\varDelta KE_{rot}\)

Parallel Axis Theorem and Rotational Dynamics

Usually, it is common to do the math while rotating bodies about an axis passing through their centers of masses. However, it might not always be the case. We have seen previously how to find the moment of inertia while being rotated through different axes, but now, we will see how we can easily determine the moment of inertia of a body through an axis parallel to the one passing through the center of mass.

::: center :::

\(\overrightarrow{\mathbf{r}}_{S, d m}=\overrightarrow{\mathbf{r}}_{S, \mathrm{cm}}+\overrightarrow{\mathbf{r}}_{d m}\) \(\left|\overrightarrow{\mathbf{r}}_{\mathrm{cm}, \perp, d m}\right|=r_{\perp, d m}\) \(\left|\overrightarrow{\mathbf{r}}_{S, \perp, d m}\right|=r_{S, \perp, d m}\) \(\left|\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}}\right|=d_{S, \mathrm{cm}}\) \(\begin{array}{l} \overrightarrow{\mathbf{r}}_{S, \perp, d m}=\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}}+\overrightarrow{\mathbf{r}}_{\perp, d m} \\ \overrightarrow{\mathbf{r}}_{S, \|, d m}=\overrightarrow{\mathbf{r}}_{S, \|, \mathrm{cm}}+\overrightarrow{\mathbf{r}}_{\|, d m} \end{array}\) \(I_{S}=\int_{\text {body }} d m\left(r_{S, \perp, d m}\right)^{2}\) \(\begin{aligned} \left(r_{S, \perp, d m}\right)^{2} &=\overrightarrow{\mathbf{r}}_{S, \perp, d m} \cdot \overrightarrow{\mathbf{r}}_{S, \perp, d m} \\ &=\left(\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}}+\overrightarrow{\mathbf{r}}_{\perp, d m}\right) \cdot\left(\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}}+\overrightarrow{\mathbf{r}}_{\perp, d m}\right) \\ &=d_{S, \mathrm{cm}}^{2}+\left(r_{\perp, d m}\right)^{2}+2 \overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}} \cdot \overrightarrow{\mathbf{r}}_{\perp, d m} \end{aligned}\) \(I_{S}=\int_{\text {body }} d m d_{S, \mathrm{cm}}^{2}+\int_{\text {body }} d m\left(r_{\perp, d m}\right)^{2}+2 \int_{\text {body }} d m\left(\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}} \cdot \overrightarrow{\mathbf{r}}_{\perp, d m}\right)\) \(d_{S, \mathrm{cm}}^{2} \int_{\mathrm{body}} d m=m d_{S, \mathrm{cm}}^{2}\) \(I_{\mathrm{cm}}=\int_{\mathrm{body}} d m\left(r_{\perp, d m}\right)^{2}\) \(2 \int_{\text {body }} d m\left(\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}} \cdot \overrightarrow{\mathbf{r}}_{\perp, d m}\right)=\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}} \cdot 2 \int_{\text {body }} d m \overrightarrow{\mathbf{r}}_{\perp, d m}\) \(2 \int_{\text {body }} d m\left(\overrightarrow{\mathbf{r}}_{S, \perp, \mathrm{cm}} \cdot \overrightarrow{\mathbf{r}}_{\perp, d m}\right)=0\) \(I_{S}=I_{\mathrm{cm}}+m d_{S, \mathrm{cm}}^{2}\) This proof tells us that if we know the moment of inertia of a rigid body rotating through its center of mass, we can determine its moment of inertia through any axis parallel to the axis passing through the center of mass. For example, let’s see a rod rotating on one end instead of its center of mass. We have seen that for a rod rotating through its center of mass, its moment of inertia is given by: \(I_{cm}=\dfrac{ML^2}{12}\) If we choose its rotation axis to be one end of the rod, that makes the distance from the axis through the center of mass to the axis through one end: \(d=\dfrac{L}{2}\) Using parallel axis theorem, we can determine that the moment of inertia is: \(I_s = I_{cm}+Md^2\) \(I_s = \dfrac{ML^2}{12} + M(\dfrac{L}{2})^2\) \(I_s = \dfrac{ML^2}{3}\)

Angular Momentum

The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity( or the cross product between r and linear momentum). It is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle if there is no external torque on the object. \(\textbf{L}=\textbf{r}\times\textbf{B}\text{ or}\) \(\text{L}=\text{I}\omega\) We have seen earlier that for an object rotating with a net torque acting on it, \(\tau_{net}=I\alpha\) We also know that a net torque bi-implies angular acceleration, thus: \(\tau_{net}=I(\dfrac{\omega_f-\omega_i}{\varDelta t})\) \(\tau_{net}\varDelta t=I(\omega_f-\omega_i)\) \(\tau_{net}\varDelta t=\varDelta I\omega\) Thus, \(\tau_{net}\varDelta t=\varDelta L\) The quantity $\tau_{net}\varDelta t$ is called the angular impulse of a body and is the rotational equivalent of impulse. We see that if the net torque on a rigid body is 0, then its change in angular momentum is 0 meaning angular momentum is conserved. Conservation of angular momentum is applicable in real life in multiple places. One case is with ballet-dancers where they can change how they are dancing and that, as a result, affecting how fast they are rotating.

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An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is very small that it is negligible. In image (b), her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy.