Rotational Motion Question 20

Question: A small particle of mass m is projected at an angle $ \theta $ with the x-axis with an initial velocity v0 in the $ x-y $ plane . At a time $ t<\frac{v _0\sin \theta }{g}, $ the angular momentum of the particle is -

[AIEEE 2010]

Options:

A) $ \frac{1}{2}mgv _0t^{2}\cos \theta \hat{i} $

B) $ -mgv _0t^{2}\cos \theta \hat{j} $

C) $ mgv _0t\cos \theta \hat{k} $

D) $ -\frac{1}{2}mgv _0t^{2}\cos \theta \hat{k} $

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Answer:

Correct Answer: D

Solution:

  • [d] at any time t So, where $ \hat{i},\hat{j} $ and $ \hat{k} $ are unit vectors along $ x,y $ and z-axis respectively.

    $ \mathbf{L} = m(\mathbf{r} \times \mathbf{v}) = -\frac{1}{2} m g v_0 t^2 \cos(\theta) \mathbf{k} $

    $\mathbf{r} = v_0 \cos(\theta) t \mathbf{i} + \left(v_0 \sin(\theta) t - \frac{1}{2} g t^2\right) \mathbf{j} $

    (m) represents mass.

    $\mathbf{v} = v_0 \cos(\theta) \mathbf{i} + \left(v_0 \sin(\theta) - g t\right) \mathbf{j} $



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