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Friction and Vertical Circular Motion Problems

A particle of...

Question

A particle of mass $1 k g$ is moving along a straight line $y = x + 4$ . Both $x$ and $y$ are in metres. Velocity of the particle is $2 m / s .$ If the magnitude of angular momentum of the particle about origin is $x \sqrt{2} k g . m^{2} / s$ , find the value of $x$ .

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Solution

$r_{⊥} = 4 cos 45^{\circ} = 2 \sqrt{2} m$
$L_{0} = m v r_{⊥}$
$= (1) (2) (2 \sqrt{2})$
$= 4 \sqrt{2} k g . m^{2} / s$

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