Section9.11Rotational Impulse and Angular Momentum

Earlier we derived Newton's second law for rotation of a rigid body about a fixed axis to be

$$\mathcal{T}_{\text{net,ext}} = I\,\alpha,\label{eq-rotational-impulse-law-old-second-law}\tag{9.11.1}$$

(Calculus step) Now, let us integrate both sides over an interval of time, say from $t=t_i$ to $t=t_f\text{.}$ We will get

$$\int_{t_i}^{t_f} \mathcal{T}_{\text{net,ext}}\, dt = \int_{t_i}^{t_f} I\,\alpha\, dt.\label{eq-integrate-torque-and-I-alpha}\tag{9.11.2}$$

We recognize the left side as just the net rotational impulse. Now, on the right side, suppose we replace $\alpha$ on the right side by $d\omega/dt\text{,}$ cancel out $dt\text{,}$ and then complete the integral. We get

$$J_{\text{rot,net}} = I \omega_f - I \omega_i,\tag{9.11.3}$$

where $\omega_i$ and $\omega_f$ are angular velocities at the initial and final instants. Since $I\omega$ is angular momentum for rotation about a principal axis, find that net impulse equals change in angular momentum.

$$J_{\text{rot,net}} = \Delta L,\label{eq-rotational-impulse-law}\tag{9.11.4}$$

where

$$\Delta L = L_f - L_i = I \omega_f - I \omega_i.\label{eq-rotational-impulse-law-ang-mom-change}\tag{9.11.5}$$

Another way of arriving at result in Eq. (9.11.4) is by looking at plots. If we plot both sides of Eq. (9.11.1) versus time and compare the areas under the two curves, the left side gives the change in angular momentum and the right side gives the net rotational impulse by external forces as shown in Figure 9.11.1.

Remark9.11.2.

We have derived Eq. (9.11.4) from Eq. (9.11.1), which is, strictly speaking, only correct if we assume $I$ is constant in time. So, it would appear that we should not use Eq. (9.11.4) if $I$ is changing.

However, more general considerations in a different derivation shows that if moment of inertia varies during the time interval, say from being $I_i$ at $t_i$ to $I_f$ at $t_f\text{,}$ then, you can still use Eq. (9.11.4), but now, we will have to use a different formula for $\Delta L\text{.}$

$$\Delta L = I_f\omega_f - I_i\omega_i\tag{9.11.6}$$

The plot in Figure 9.11.4 shows net torque (with countercloclwise positive) on a disk of moment of inertia $5.0\text{ kg.m}^2$ as a function of time. At $t=0\text{,}$ the disk is rotating at angular velocity $2.0\text{ rad/s}$ in the clockwise direction. What is the magnitude and sense of rotation of the angular velocity at $t = 5.0\text{ sec}\text{?}$

Hint

Use rotational impulse/change in angular momentum.

$1.0\text{ rad/s}$ in the clockwise direction.

Solution

We will use $J_\text{rot}^\text{net} = \Delta L$ with $J_\text{rot}^\text{net} =$ area under the curve, $\Delta = (I\omega)_f - (I\omega)_i$ to find $\omega_f\text{.}$ The area above zero torque give positive impulse and that below give negative impulse.

\begin{equation*} J_\text{rot}^\text{net} = \dfrac{1}{2}\times 10 \times 3 + (-5\times 2) = 15-10 = 5\text{ N.m.s}. \end{equation*}

Now, note that $\omega_i = -2\text{ rad/s}$ since it is clockwise and we are using counterclockwise sense as positive. Therefore, rotational impulse/change in angular momentum gives

\begin{equation*} 5\,\omega_f - 5 \times (-2) - 5. \end{equation*}

Therefore,

\begin{equation*} \omega_f = -1.0\text{ rad/s}. \end{equation*}

Therefore, the final angular velocity is $1.0\text{ rad/s}$ in the clockwise direction.

Subsection9.11.1(Calculus) Change in Angular Momentum by a Rotational Impulse

From the general second law of motion,

\begin{equation*} \vec F =\dfrac{d\vec p}{dt}, \end{equation*}

we derived before that

\begin{equation*} \dfrac{ d\vec L }{dt} = \vec{\mathcal{T}}. \end{equation*}

Multiply both sides by $dt \text{,}$ we can integrate this from $t= t_i$ to $t= t_f \text{.}$

\begin{equation*} \int_{t_i}^{t_f}\vec{\mathcal{T}}\,dt = \vec L_f - \vec L_i, \end{equation*}

which is the statment that net external rotational impulse equals the change in angular momentum.

$$\vec{J}_\text{rot} = \vec L_f - \vec L_i.\label{eq-rotational-impulse-law-calc}\tag{9.11.7}$$