The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $ heta$ and then released. The bob will then pass through its equilibrium position with a speed $v,$ where $v$ equals

Question

The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $	heta$ and then released. The bob will then pass through its equilibrium position with a speed $v,$ where $v$ equals

Dr. Abubacker · Accepted Answer

(a) When the bob of pendulum is brought to a position making an angle $	heta$ with the equilibrium position, then height of fall of pendulum will be,
$h = l - l\cos	heta = l(1 - \cos	heta)$.
Taking free fall of the
$u = 0,\ a = g,\ g = h = l\left( 1 - \cos	heta ight),\ v = ?$
Now, $v^{2} = u^{2} + 2gh = 0 + 2gl\left( 1 - \cos	heta ight)$
or $v = \sqrt{2gl(1 - \cos	heta)}$

Q. The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed $v,$ where $v$ equals

Explanation

Reviewed by Dr. Abubacker

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