Question

3. The Centripetal Acceleration of the bob of a conical pendulum is
a) rg/coso b) rg/1 c) g/ dy rg/l coso
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Answer 1

Explanation:

The acceleration is a centripetal acceleration, which is ω2 r. ω is the angular velocity, which is equal to the tangential velocity (v) divided by the radius (r)...

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Answer 2

Given:

A Bob is undergoing motion in conical pendulum

To find:

Centripetal acceleration of the bob?

Calculation:

Let the mass of the bob be "m" , its instantaneous radius of rotation be "r" and length of string be "l".

Considering the tension in the string to be T, it can be broken down into perpendicular components as follows :

$\sf \: 1) \: T \sin( \theta) = \dfrac{m {v}^{2} }{r} \\ \sf \: 2) \: T \cos( \theta) = mg$

Dividing the 2 Equations:

$\sf \therefore \: \dfrac{ \sin( \theta) }{ \cos( \theta) } = \dfrac{ {v}^{2} }{rg}$

$\sf \implies \: \tan( \theta) = \dfrac{ (\frac{{v}^{2}}{r}) }{g}$

$\sf \implies \: \tan( \theta) = \dfrac{a_{c} }{g}$

$\sf \implies \: a_{c} = g \tan( \theta)$

$\sf \implies \: a_{c} = g \times \dfrac{r}{ \sqrt{ {l}^{2} - {r}^{2} } }$

$\sf \implies \: a_{c} = \dfrac{gr}{ \sqrt{ {l}^{2} - {r}^{2} } }$

So, final answer is :

$\boxed{ \bf \: a_{c} = \dfrac{gr}{ \sqrt{ {l}^{2} - {r}^{2} } } }$