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A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)?
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A particle of mass m is projected with a velocity v making an angle of...
Solution:


Definition of angular momentum:

Angular momentum is defined as the product of moment of inertia and angular velocity of a body about an axis of rotation.

Equation for angular momentum:

L = Iω

where,
L = Angular momentum
I = Moment of inertia
ω = Angular velocity

Calculation of angular momentum:

At the highest point, the vertical component of velocity becomes zero. So, the projectile moves only in the horizontal direction. Hence, the moment of inertia about the point of projection is given by:

I = mr²

where,
m = Mass of the particle
r = Distance of particle from point of projection

At the highest point, the height of the particle is given by:

h = (v²sin²θ)/2g

where,
θ = Angle of projection
v = Initial velocity of the particle
g = Acceleration due to gravity

Substituting the values in the above equation, we get:

h = (v²/2g)

Therefore, the distance of the particle from the point of projection is given by:

r = h/√2

Substituting the values, we get:

r = v²/2g√2

The angular velocity of the particle is given by:

ω = v/r

Substituting the values, we get:

ω = √2g/v

Therefore, the angular momentum of the particle is given by:

L = Iω = (mr²)(√2g/v)

Substituting the values, we get:

L = (m/2)(v²/2g)(2g/v)

L = mv²/4

Therefore, the magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is mv²/4.

Answer:

Option (B) mv³/4√2g is incorrect. The correct answer is (A) Zero. This is because at the maximum height, the particle has zero vertical velocity, and hence, zero angular momentum about the point of projection.
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A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)?
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A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)?.
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