Question

Derive an expression for the angular speed of the bob of a conical

pendulum​

Answer 1

Answer:

Consider a conical pendulum of string length L with its bob of mass m performing UCM along a circular path of radius r.

At every instant of its motion, the bob is acted upon its weight mg and the tension F in the string. If the constant angular speed of the bob is ω, the necessary horizontal centripetal force if $F_{c}$=mω²r

$F_{c}$ is the resultant of the tension in the string and the weight. Resolve F into components F cos θ vertically opposite to the weight of the bob and F sin θ horizontal. F cos θ balances the weight. F sin θ is the necessary centripetal force.

∴ F sin θ = mω²r ... (1)

and F cos θ = mg ... (2)

(check the diagram)

Conical Pendulum

Dividing Eq. (1) by Eq. (2),

tan θ = ω²r/g

From the diagram,

tan θ = $\frac{r}{OC}$

tan θ = $\frac{r}{h}$

tan θ = r/L cos θ ... (3)

$\frac{r}{h}$ =  ω²r/g

∴ ω² = $\frac{g}{h}$

∴ ω² = g/L cos θ ... (4)

The angular speed of the bob,

ω = $\sqrt{\frac{g}{h}$

ω = $\sqrt{\frac{g}{L cos theta}$ ... (5)

is the required expression for ω.

[Note: From Eq. (4), cos θ= g/ω²L. Therefore, as ω increases, cos θ decreases and θ increases.]

If n is the frequency of revolution of the bob,

ω = 2nπ

ω = $\sqrt{g/L cos theta}$

∴ n = 1/2(π)$\sqrt{g/LcosTheta}$ ... (6)

is the required expression for frequency.

Hence, the expressions are derived.

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