Question

QUESTION ❓✔️❓

Q 1) What is a conical pendulum?
Show that its time period is given by,
$2\pi \sqrt{ \frac{l \cos(θ) }{g} }$
where I is the length of the string,
θ is the angle that the string makes with the vertical and g is the acceleration due to gravity.

QUESTION ❓✔️❓

Q. 2) Drive an expression for centripetal acceleration of a particle performing uniform circular motion​

Answer 1

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Conical pendulum

A simple pendulum, which is given such a motion that the bob describes a horizontal circle and the string making a constant angle with the vertical describes a cone, is called a conical pendulum

(2)

The attached figure shows the free body diagram for a conical pendulum.

The bob has no vertical displacement.

Hence, Tcosθ=mg

Tsinθ = mv²/r

Eliminating T, we get:

tanθ= v²/rg

$v = \sqrt{rg \: tanθ} \\ v = \sqrt{rg \times \frac{r}{h} } \\ v = \sqrt{ \frac{r {}^{2} gl}{hl} } \\ v = \sqrt{ \frac{r {}^{2} g}{l \cos(θ) } }$

Time period of conical pendulum is:

$T = \frac{2 \pi \: r}{v} \\ T = \frac{2\pi \sqrt{l \cos(θ) } }{ \sqrt{p} } \\ T = \frac{4\pi \sqrt{l \cos(θ) } }{ \sqrt{4g} }$

Centripetal acceleration of a particle performing uniform circular motion is given by a꜀=v²/r=ω²r

where v=linear velocity,r=radius and ω=angular velocity.

Answer 2

Answer:

water which collects as droplets on a cold surface when humid air is in contact with it.

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