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Q.

A particle is moving in a circular path of radius a with a constant velocity v as shown in the figure. The center of circle is marked by ‘C’. The angular momentum from the origin O can be written as

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a

$mvacos 2 θ$

b

$mva (1 + \cos 2 θ)$

c

$mva (1 + \cos θ)$

d

mva

answer is A.

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Detailed Solution

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$In ∆ OPC,$

$\frac{x}{\sin (π - 2 θ)} = \frac{a}{sinθ}$

$\Rightarrow x = 2 acosθ$

$| \vec{L} | = mvx sinϕ here, ϕ = 90 ° - θ$

$\Rightarrow | \vec{L} | = mv (2 acosθ) cosθ = mva [1 + \cos 2 θ]$

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