A particle of mass m = 5 units is moving with uniform speed V = 3 2 units in the X Y plane along the line Y = X + 4 . The magnitude of the angular momentum about origin is

Angular momentum = momentum × perpendicular distance. L = m v b cos ⁡ θ = 5 × 3 2 × 4 × cos ⁡ 45 o L = 60 kg m 2 s - 1

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationAngular momentumEasy2 minPYQ_2020

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

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Answer:

Solution:

Angular momentum = momentum × perpendicular distance.

$L = m v b \cos θ$

$= 5 \times 3 \sqrt{2} \times 4 \times \cos 45^{o}$

$L = 60 kg m^{2} s^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:RotationSubtopic:Angular momentum

⏱ 2mℹ️ Source: PYQ_2020

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