JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Rotational Mechanics

A particle of mass 100 g is projected at time

t = 0

with a speed

20 ms^{- 1}

at an angle

4 5^{\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 = 2 s

is found to be

K kg m^{2} / s

. The value of K is ______.
(Take

g = 10 ms^{- 2}

)

Answer: 800

Solution

Taking the launch point as origin, the only torque about it comes from gravity,

τ = m g x

, where

x = (v cos 4 5^{\circ}) t

is the horizontal distance. The angular momentum built up is

L = \int_{0}^{t} m g (v cos 4 5^{\circ}) t^{'} d t^{'} = m g (v cos 4 5^{\circ}) \frac{t ^{2}}{2}

. With

m = 0.1

kg,

v cos 4 5^{\circ} = 102

m/s,

g = 10

m/s

^{2}

and

t = 2

L = 0.1 \times 10 \times 102 \times \frac{4}{2} = 202 = 800

kg m

^{2}

/s. The value of

K

800

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer