A particle of mass 100 g is projected at time t = 0 with a speed 20 m s – 1 at an angle 45 ° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the s

Question

A particle of mass 100 g is projected at time t = 0 with a speed 20 m s – 1 at an angle 45 ° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2 s is found to be K kg m 2 s - 1 . The value of K is ______. (Take g = 10 m s - 2 )

TestHub · Accepted Answer

As torque due to gravitational force will be variable, therefore we can use

$Δ L = \int_{0}^{t} τ d t$ .

Now, torque of $m g$ at time $t$ $= (v_{x} t) m g$ and horizontal velocity of the projectile will remain constant $(v_{x} = v \cos 45 ° = 20 \times \frac{1}{\sqrt{2}} = 10 \sqrt{2})$ .

$L_{0} = \int_{0}^{2} m g (v_{x} t) d t$

$= m g v_{x} \frac{t^{2}}{2} = (0.1) (10) (10 \sqrt{2}) \frac{2^{2}}{2}$

$= 20 \sqrt{2}$

$= \sqrt{800} kg m^{2} s^{- 1}$

Therefore, $K = 800$ .

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