Rotational Inertia and Kinematics

Given \(m_i\), mass, \(\vec{r_i}'\), location of the center of mass, \(l_i\), \(\omega\), the angular velocity, figure a \(KE_{tot,rot}\).

Because of the fact that the value \(\omega\) is in units \(\frac{d\theta}{dt}\), the rate of radians change, and we know of a radius of the spin \(l_i\), we could figure the velocity at which it is moving by simply scaling the change in radians up to a circle of radius \(l_i\), that is:

\begin{equation} V_i' = l_i \omega \end{equation}

(note that, to understand this, radians \(\frac{arc length}{radius}\))

And so, substituting into the statement of \(\sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2\)

\begin{align} KE_{rot} =& \sum^N_{i=1} \frac{1}{2}m_i\vec{v_i}'^2 \\ =& \sum^N_{i=1} \frac{1}{2}m_i(l_i \omega)^2 \\ =& \sum^N_{i=1} \frac{1}{2}m_i l_i^2 \omega^2 \\ =& \frac{1}{2}\omega^2 \sum^N_{i=1} (m_i l_i^2) \end{align}

We begin by defining a coordinate system such that positive values are pointed "downwards". That is, as values increase in the positive direction, their corresponding vectors are pointed further towards the "down" direction.

Given \(a=a_0\), initial velocity \(v_0\), and position \(y_0\), we derive the kinematics equations.

\begin{align} a(t) =& a_0 \\ \int a(t) dt =& \int a_0 dt \\ v(t) =& a_0t + C \end{align}

We are given that \(v(0)=v_0\). \(v(0) = C = v_0\), hence, \(C=v_0\). The velocity statement is therefore,

\begin{equation} v(t) = a_0t+v_0 \end{equation}

Continuing with integration:

\begin{align} v(t) =& a_0t + v_0 \\ \int v(t) =& \int a_0t + v_0 dt \\ y(t) =& \frac{1}{2}a_0t^2+v_0t+C \\ \end{align}

Again, substituting \(C = y_0\) by the same logic above — \(y(0) = C = y_0\), we derive the statement for the position equation.

\begin{equation} y(t) = \frac{1}{2}a_0t^2 + v_0t + y_0 \end{equation}

But what if we want to do this in multiple dimensions? They will be vectors, for one.

\begin{align} \vec{a}(t) =& \vec{a_0} \\ \int\vec{a}(t)dt =& \int\vec{a_0} dt \end{align}

and so on. The reason why we could do this is simply because \(\vec{a_0} = a_0_x \hat{i} + a_0_y\hat{j} + a_0_z\hat{k}\), and, perhaps unsurprisingly, integrals and derivatives and commutative and distributive across addition.

The Kinematics Equations derivations above relied on the fact that the coordinate system was defined as "positive downwards". Were this not to be the case, constants would have to be redefined and the signs of most terms would be flipped:

Given \(a=-a_0\), initial velocity \(-v_0\), and position \(-y_0\), we (re)derive the kinematics equations.

\begin{align} a(t) =& -a_0 \\ \int a(t) dt =& \int -a_0 dt \\ v(t) =& -a_0t + C \end{align}

We are given that \(v(0)=-v_0\). \(v(0) = C = -v_0\), hence, \(C=-v_0\). The velocity statement is therefore,

\begin{equation} v(t) = -a_0t-v_0 \end{equation}

Continuing with integration:

\begin{align} v(t) =& -a_0t - v_0 \\ \int v(t) =& \int -a_0t - v_0 dt \\ y(t) =& \frac{-1}{2}a_0t^2 - v_0t+C \\ \end{align}

Again, substituting \(C = -y_0\) by the same logic above — \(y(0) = C = -y_0\), we derive the statement for the position equation.

\begin{align} y(t) =& \frac{-1}{2}a_0t^2 - v_0t - y_0 \\ y(t) =& -(\frac{1}{2}a_0t^2 + v_0t + y_0) \end{align}

As such, if the signage were flipped, terms of the kinematics equation would therefore be the negative of the original — that \(y'(t) = -y(t)\) if \(y'\) contains uses a flipped coordinate system. Leveraging the same property, therefore, we could derive quickly the other two statements with a flipped coordinate system:

\begin{align} \Delta y' =& \frac{-v(t_1)-v(t_2)}{2}\Delta t \\ =& -\frac{v(t_1)+v(t_2)}{2}\Delta t \end{align}

and

\begin{align} v^2(t) =& {-v_0}^2 + 2a_0 (-y(t) - (-y_0)) \\ =& {v_0}^2 + 2a_0 (y_0 - y(t)) \end{align}

\begin{equation} v^2(t) \stackrel{?}{=} {v_0}^2 + 2a_0(y(t)-y_0) \end{equation}

We begin by deriving an equation of \(a(y)\) — acceleration as a function of position.

Based on first principles, the following is true:

\begin{equation} a(t) = \frac{dv}{dt} \end{equation}

With apply the chain rule, we derive the following:

\begin{align} a(t) =& \frac{dv}{dt} \\ \Rightarrow \frac{dv}{dt} =& \frac{dv}{dy}\frac{dy}{dt} \\ \end{align}

Integrating both sides w.r.t. position…

\begin{align} \frac{dv}{dt} =& \frac{dv}{dy}\frac{dy}{dt} \\ \Rightarrow \int \frac{dv}{dt}\ dy =& \int \frac{dv}{dy}\frac{dy}{dt}\ dy\\ \end{align}

The right integral now requires some additional movement to solve.

\begin{align} \int \frac{dv}{dt}v\ dy =& v^2 - \int \frac{dv}{dy} v\ dy \\ \Rightarrow 2\int \frac{dv}{dy}v\ dy =& v^2 + C \\ \Rightarrow \int \frac{dv}{dy}v\ dy =& \frac{v^2}{2} + C \end{align}

Substituting the right statement back into the original expression, and continuing to solve.

\begin{align} & \int \frac{dv}{dt}\ dy = \frac{v^2}{2} + C \\ & \Rightarrow \int \frac{dv}{dt}\ dy = \frac{v^2}{2} + C\\ & \Rightarrow 2ay + C = v^2 \end{align}

Because of the fact that this expression is a value of \(v^2\), and that the constant is an offset of the value (as per calculated by \(2ay\)), we set \(C = {v_0}^2\). Additionally, \(y\) is an statement that represents the offset of \(y\) position, we set \(y = y(t)-y_0\) to represent the actual offset.

Hence:

\begin{equation} v^2 = 2a(y(t)-y_0) + {v_0}^2 \end{equation}