A particle of mass = 5 is moving with a uniform speed $v = 3\sqrt{2}$ in the $XOY$ plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is

Question

Dr. Abubacker · Accepted Answer

(a) Motion is along the time ; $Y = X + 4;$
Differentiating it wrt time, we have
$\frac{dY}{dt} = \frac{dX}{dt}ie,v_{Y} = v_{X}$
As, $v = \left( v_{X}^{2} + v_{Y}^{2} ight)^{1/2} = 3\sqrt{2}$ and $v_{X} = v_{Y},$ therefore,
$\left( v_{X}^{2} + v_{X}^{2} ight)^{1/2} = 3\sqrt{2}$ or $v_{X} = 3 = v_{Y}$
When $X = 0,$ from the given equation,
$Y = 0 + 4 = 4$
Magnitude o angular momentum of particle
$= m\ v\ r = m\ v\ y\ \ \ \ \ \ (\because y = r)$
$= 5 	imes 3 	imes 4 = 60$ units

Q. A particle of mass = 5 is moving with a uniform speed $v = 3\sqrt{2}$ in the $XOY$ plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is

Explanation

Reviewed by Dr. Abubacker

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