Question

$\vec{F} = a \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $\vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k}$ The value of ‘a’ for which the angular momentum is conserved is

Moderate

Solution

$\vec{τ} = \vec{r} \times \vec{F} and τ = \frac{dL}{dt} = 0$ i.e. $\vec{r} a n d \vec{F} a r e p a r a l l e l \frac{a}{2} = \frac{3}{- 6} = \frac{6}{- 12} a = - 1$

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