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} $
Show Answer
Answer:
Correct Answer: D
Solution:
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[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} $
BETA
