Monday, May 28, 2007

MCQ on Angular Momentum of a Projectile

A particle of mass ‘m’ is projected with a velocity ‘v’ making an angle of 45º with the horizontal. When the projectile is at its maximum height, the magnitude of its angular momentum about an axis passing through the point of projection and perpendicular to the plane of its path is

(a) zero (b) mv2/4√2g (c) ) mv3/4√2g (d) mv2/√2g (e) mv3/√2g

You will find questions similar to this on many occasions. So, take a special note of this question.

The velocity of a projectile changes continuously along its path because of the change in the vertical component of velocity under the gravitational pull. If θ is the angle of projection, the horizontal component of velocity, vcosθ remains unchanged throughout the path and at the maximum height, the vertical component of velocity is zero and it has the horizontal velocity vcosθ only. The ‘lever arm’ for angular momentum at the maximum height is the maximum height (v2sin2θ)/2g itself so that the angular momentum is (mvcosθ)×(v2sin2θ)/2g = (mvcos45º )×(v2sin2 45º )/2g = mv3/4√2g

A simplified form of this question appeared in KEAM (Medical) 2007 question paper:

A particle is projected with a speed ‘v’ at 45º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the projectile is at its maximum height ‘h’ is

(a) zero (b) mvh2/√2 (c) mv2h/2 (d) mvh3/√2 (e) mvh/√2

Since the maximum height is given as ‘h’, you can write the answer in no time as (mvcos45º)×h = mvh/√2.

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