A body of mass 5 kg moving with a uniform speed $3 \sqrt{2} \mathrm{~ms}^{-1}$ in $\mathrm{X}-\mathrm{Y}$ plane along the line $\mathrm{y}=\mathrm{x}+4$. The angular momentum of the particle about the origin will be ______ $\mathrm{kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Rotational Mechanics

A body of mass 5 kg moving with a uniform speed

32 ms^{- 1}

X - Y

plane along the line

y = x + 4

. The angular momentum of the particle about the origin will be ______

kg m^{2} s^{- 1}

Answer: 60

Solution

Angular momentum about the origin is

L = m v d

, where

d

is the perpendicular distance from the origin to the line of motion.

The line of motion is

y = x + 4

, or

x - y + 4 = 0

. The perpendicular distance from the origin is

d = \frac{∣4∣}{1 ^{2} + 1 ^{2}} = \frac{4}{2} = 22 m

L = m v d = 5 \times 32 \times 22 = 5 \times 3 \times 2 \times 2 = 60 kg m^{2} s^{- 1}

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer

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