Q7 E - Fundamentals Of Differential Equations And Boundary Value Problems

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 gives torque $=-\ell \mathrm{mgsin\theta }$.

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

Step-by-Step Solution

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 1: 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.

 

Step 2: 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.

 

Step 3: 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,  $\mathrm{T}=\mathrm{lF}$  .

 

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.

Step 4: 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