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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Rotational Motion

BITSAT 2013 · Physics · Rotational Motion

A particle of mass ' m ' is projected with a velocity v making an...

A particle of mass ' mm ' is projected with a velocity vv making an angle of 30∘30^{{}^{\circ}} with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' hh ' isA particle of mass ' m ' is projected with a velocity v making an angle of 30^ ^ with the horizontal. The magnitude of

Options

  1. A32mv2g\frac{\sqrt{3}}{2} \frac{m v^{2}}{g}
  2. Bzero
  3. Cmv32g\frac{m v^{3}}{\sqrt{2 g}}
  4. D316mv3g\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}

Correct answer

D. 316mv3g\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}

Step-by-step solution

Angular momentum L0=pr⊥(∵L_{0}=p r_{\perp} \left(\because\right.linear momentum p=mvcos⁡θandr1=H)p=m v \cos \theta and \left.\mathrm{r}_{1}=H\right) A particle of mass ' m ' is projected with a velocity v making an angle of 30^ ^ with the horizontal. The magnitude of
⇒L0=mvcos⁡\thetaH=mv32⋅v2sin⁡230∘2g=3mv316g\Rightarrow L_{0}=m v \cos \thetaH=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{{}^{\circ}}}{2 g}=\frac{\sqrt{3} m v^{3}}{16 g}
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