Question

Use the result of Problem 8 to find the value of $\mathrm{\theta }\text{'}\left(0\right)$, the initial velocity, that must be imparted to a pendulum at rest to make it approach (but not cross over) the apex of its motion. Take l = g for simplicity.

Short answer

Therefore, the value of the initial velocity is $\mathrm{\theta }\text{'}\left(0\right)=\pm 2$.

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 is $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$constant; i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Change of angular momentum:

$\mathrm{m}{\ell }^2\mathrm{\theta }=-\ell \mathrm{mgsin\theta }$…… (1)

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

Prove the given equation.

Referring to Problem 8:$\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta }=\mathrm{constant}$ …… (2)

To find the value of $\mathrm{\theta }\text{'}\left(0\right)$.

Given, l = g.

Let us take constant = k. Then,

$\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{cos\theta }=\mathrm{k}$…… (3)

Let’s say$\mathrm{t}=0$and$\mathrm{\theta }$must be zero with some initial velocity,$\mathrm{\theta }\text{'}\left(0\right)={\mathrm{\theta }}_0$and $\mathrm{t}=\mathrm{\pi }$the velocity of the pendulum must be zero. So,$\mathrm{\theta }\text{'}\left(\mathrm{\pi }\right)=0$.

Now, implement the conditions.

Put $\mathrm{t}=0$in equation (3).

role="math" localid="1664029463502" $$\begin{gathered} \frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{cos\theta }=\mathrm{k} \\ \frac{1}{2}\mathrm{\theta }\text{'}{\left(0\right)}^2-\cos \left(0\right)=\mathrm{k} \\ \frac{1}{2}{\mathrm{\theta }}_0^{\text{'}2}-1=\mathrm{k} \end{gathered}$$

Now, put $\mathrm{t}=\mathrm{\pi }$.

$$\begin{gathered} \frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{\pi }\right)}^2-\cos \left(\mathrm{\pi }\right)=\mathrm{k} \\ \mathrm{k}=1 \end{gathered}$$

Then,

$$\begin{gathered} \frac{1}{2}{\mathrm{\theta }}_0^{\text{'}2}-1=\mathrm{k} \\ {\mathrm{\theta }}_0^{\text{'}2}=4 \\ {\mathrm{\theta }}_0=\pm 2 \end{gathered}$$

So, the initial velocity is $\mathrm{\theta }\text{'}\left(0\right)=\pm 2$.