JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 18)
A particle of mass 100 g is projected at time t = 0 with a speed 20 ms$$^{-1}$$ at an angle 45$$^\circ$$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2s is found to be $$\mathrm{\sqrt K~kg~m^2/s}$$. The value of K is ___________.
(Take g = 10 ms$$^{-2}$$)
_29th_January_Evening_Shift_en_18_1.png)
Explanation
Horizontal displacement $$x = v\cos \theta t$$
$$ = 10\sqrt 2 t$$
So torque of weight about point of projection is
$$\tau = mgx\,.\,( - \widehat k)$$
$${{d\overrightarrow L } \over {dt}} = mgx( - \widehat k)$$
$$\int_0^{\overrightarrow L } {d\overrightarrow L = 0.1 \times 10 \times 10\sqrt 2 \int_0^2 {t\,dt( - \widehat k)} } $$
$$\overrightarrow L = - 20\sqrt 2 \widehat k$$
$$|\overrightarrow L | = 20\sqrt 2 = \sqrt {800} $$ kg m$$^2$$/s
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