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Section 9.10 Rotational Impulse

Recall that when a force acts on a body, we define the influence of the force by impulse. Impulse of a force causes change in momentum of a body. Similarly, we define the influence of a torque by rotational impulse. We will prove in a section below that the rotational impulse causes change in angular momentum of the body.

Rotational Impulse of a Constant Torque:

If a constant torque \(\mathcal{T}\) acts for a duration \(\Delta t\text{,}\) the rotational impulse, \(J_{\text{rot}}\text{,}\) of the torque is gven by the product of torque and duration.

\begin{equation} J_{\text{rot}} = \mathcal{T}\, \Delta t\ \ \ (\text{constant torque}) \tag{9.10.1} \end{equation}

Since torque is a vector, \(J_{\text{rot}}\) is also a vector. Similar to torque, we can give the direction of rotational impulse either by the clockwise/counterclockwise sense of rotation or by its direction being towards the axis of rotation or opposite to the axis.

Consider a wheel which is struck tangentially by a constant force at the edge as shown in Figure 9.10.1. Since the lever arm of this force is the radius \(R \text{,}\) the magnitude of the torque is \(RF\text{.}\)

Rotational impulse of a this torque over a duration \(\Delta t\) will just be the product of torque and duration.

\begin{equation*} J_{\text{rot}} = R F \Delta t. \end{equation*}

Figure 9.10.1.

The situation is a bit more complicated if either the magnitude or the direction of the torque is changing. In this chapter, we are mostly interested in fixed-axis rotations, therefore, we will examine circumstances in whcih direction of torque is fixed but may vary in magnitude with time.

\begin{equation*} \end{equation*}

Rotational Impulse of a Varying Torque:

In Figure 9.10.1, force \(F\) was continuously actimg for duration \(\Delta t\) with same magnitude. But, sometimes, we have a situation force may act for a short period of time with varying magnitude, e.g., when you strike wheel by a hammer and make the wheel rotate as in Figure 9.10.2.

Figure 9.10.2.

The torque by hammer can be approximated to have a rising and falling shape as shown in Figure 9.10.3. Now, to get the rotational impulse from \(\mathcal{T}(t)\) we need to integrate the torque, i.e.,

\begin{equation*} J_{\text{rot}} = \int \mathcal{T}(t)\, d t. \end{equation*}
Figure 9.10.3.

The result can also be obtained by computing the area under the curve of the plot of \(\mathcal{T} \) versus \(t \text{.}\)

\begin{equation} J_{\text{rot}} = \text{ Area under }\mathcal{T}\text{ versus }t. \ \ (\text{fixed direction}) \tag{9.10.2} \end{equation}

Often the lever arm \(r_{\perp} \) of a force will not be changing, then, magnitude of the rotational impulse is just the product of the lever arm and the regular impulse.

\begin{equation*} J_{\text{rot}} = r_{\perp}\times [\text{Area under }F\text{ versus }t], \end{equation*}

which is

\begin{equation} J_{\text{rot}} = r_{\perp}\ J\ \ \ (\text{fixed direction, fixed }r_{\perp}) \tag{9.10.3} \end{equation}

Net Rotational Impulse of Multiple Forces: Since rotational impulse is a vector, we would need to add them vectorially to get the net. In a fixed-axis situation, we keep track of counterclockwise and clockwise sense of each torque, and add or subtract them accordingly to obtain the net rotational impulse.

A pulley has a rope wound around its edge as shown in the left panel of Figure 9.10.5. A boy gives a jerk to the rope to get the pulley rotating. If the force applied by the boy varies in time as shown in the right panel of Figure 9.10.5 and the radius of the pulley is \(0.5 \text{ m}\text{,}\) what is the rotational impulse of the force by the boy?

Figure 9.10.5. Figure for Checkpoint 9.10.4.
Hint

The lever arm is fixed, so use lever arm times the regular impulse.

Answer

\(3.5\text{ N.m.s} \) into-the-page.

Solution

Here, the lever arm of the force is fixed in time to \(r_{\perp} = R = 0.5\text{ m}\) and the magnitude of the regular impulse is just the area under the force-time plot. Therefore, we get

\begin{equation*} J_{\text{rot}} = r_\perp\, J = 0.5 \times \dfrac{1}{2} \times 0.200\times 70 = 3.5\text{ N.m.s}. \end{equation*}

The direction of the rotational impulse is into-the-page.

Subsection 9.10.1 (Calculus) Fundamental Definition of Rotational Impulse

Rotational impulse is the integral of torque with respect to time over the duration \(t_i \) to \(t_f\text{.}\)

\begin{equation} \vec{J}_{\text{rot}} = \int_{t_i}^{t_f}\vec{\mathcal{T}}(t)\, dt.\tag{9.10.4} \end{equation}

For a torque that has only the \(z \) component nonzero, only the \(z \) component of the rotational impulse will be nonzero.

\begin{equation} J_{\text{rot},z} = \int_{t_i}^{t_f} \mathcal{T}_z(t)\, dt.\tag{9.10.5} \end{equation}