Chapter 4: Q8 E (page 211)

Use the energy integral lemma to show that pendulum motion obeys $\frac{{(θ')}^{2}}{2} - \frac{g}{l} cosθ = constant$

Short Answer

Expert verified

Therefore, the given statement is true. The pendulum motion obeys $\frac{{(θ')}^{2}}{2} - \frac{g}{l} cosθ = constant$ .

Step by step solution

General form

The Energy Integral Lemma:

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

Change of angular momentum:

$m ℓ^{2} θ = - ℓ mgsinθ$ …… (1)

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

Prove the given equation

Referring to Problem 7: ${ml}^{2} \frac{d^{2} θ}{{dt}^{2}} = - lmgsinθ$ …… (2)

To prove: $\frac{{(θ')}^{2}}{2} - \frac{g}{l} cosθ = constant$

Let us take equation (2) to get,

${ml}^{2} \frac{d^{2} θ}{{dt}^{2}} = - lmgsinθ θ = - \frac{lmgsinθ}{{ml}^{2}} = - \frac{g}{l} sinθ$

So,

f (θ) = - \frac{g}{l} sinθ

Then,

$f (θ) = \frac{d}{dθ} F (θ) F (θ) = \int f (θ) dθ = \frac{g}{l} cosθ$

Substitute the values in E(t).

$E (t) = \frac{1}{2} θ' {(t)}^{2} - F (θ (t)) = \frac{1}{2} θ' {(t)}^{2} - \frac{g}{l} cosθ$

Hence proved.

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Use the energy integral lemma to show that pendulum motion obeys $\frac{{(θ')}^{2}}{2} - \frac{g}{l} cosθ = constant$

Short Answer

Step by step solution

General form

Prove the given equation

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Use the energy integral lemma to show that pendulum motion obeys $\frac{{(θ')}^{2}}{2} - \frac{g}{l} cosθ = constant$