Question

Pendulum Equation. To derive the pendulum equation (21), complete the following steps.

a. The angular momentum of the pendulum mass m measured about the support O in Figure 4.18 on page 208 is given by the product of the “lever arm” length $\ell$ and the component of the vector momentum mv perpendicular to the lever arm. Show that this gives angular momentum $=\mathrm{m}{\ell }^2\frac{\mathrm{d\theta }}{\mathrm{dt}}$.

b. The torque produced by gravity equals the product of the lever arm length and the component of gravitational (vector) force mg perpendicular to the lever arm. Show that this givestorque $=-\ell \mathrm{mgsin\theta }$.

c. Now use Newton’s law of rotational motion to deduce the pendulum equation (20).

1. 

Short answer

1. Therefore, the given statement is true. That is, angular momentum$=\mathrm{m}{\ell }^2\frac{\mathrm{d\theta }}{\mathrm{dt}}$is true.
2. Hence, the given statement is true. That is, torque$=-\ell \mathrm{mgsin\theta }$ is true.
3. So, the given statement is true. Equation (20) and Newton’s rotation law both are equal.

Step-by-step solution

General form 

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation$\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral of$\mathrm{f}\left(\mathrm{y}\right)$ , that is,$\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$ . Then the quantity $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant; i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Change of angular momentum:

data-custom-editor="chemistry" $\mathrm{m}{\ell }^2\mathrm{\theta }=-\ell \mathrm{mgsin\theta }.......\left(1\right)$

…… (1)

Newton’s rotational law: The rate of change of angular momentum is equal to torque.

Prove the given equation.

Given that, the figure of the pendulum is shown below.

To prove: angular momentum $=\mathrm{m}{\ell }^2\frac{\mathrm{d\theta }}{\mathrm{dt}}$…… (2)

From the figure, we can see that the angular velocity is,$\mathrm{v}=\frac{\mathrm{d\theta }}{\mathrm{dt}}$ .

Then, find the angular momentum.

$$\begin{gathered} \mathrm{L}=\mathrm{lmv} \\ =\mathrm{lm}\times \mathrm{l}\times \frac{\mathrm{d\theta }}{\mathrm{dt}} \\ ={\mathrm{ml}}^2\frac{\mathrm{d\theta }}{\mathrm{dt}} \end{gathered}$$

So, angular momentum$=\mathrm{m}{\ell }^2\frac{\mathrm{d\theta }}{\mathrm{dt}}$ is true.

Prove the statement

To prove: Torque$=-\ell \mathrm{mgsin\theta }$.

From the given figure we can see that the restoring force is,$\mathrm{F}=\mathrm{mgsin\theta }$ .

Then, the relation between torque T and force F is,role="math" localid="1663957923230" $\mathrm{T}=\mathrm{lF}$ .

Substitute the value of force in torque.

$\mathrm{T}=\mathrm{lmgsin\theta }$

Since the force and $\mathrm{\theta }$ are in opposite directions. So, $\mathrm{T}=-\mathrm{lmgsin\theta }$

Hence proved.

Prove the statement

Given that, Newton’s rotational law.

That is, the rate of change of angular momentum is equal to torque.

The angular momentum is,$\mathrm{L}={\mathrm{ml}}^2\frac{\mathrm{d\theta }}{\mathrm{dt}}$ . Then, differentiate with respect to t.

$\frac{\mathrm{dL}}{\mathrm{dt}}={\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}$

Then,

${\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\mathrm{lmgsin\theta }$

So, which is similar to equation (1). Then, the statement is true.

Hence proved