JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 26)
A particle of mass 'm' is moving in time 't' on a trajectory given by
$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$
Where $$\alpha$$ and $$\beta$$ are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.
$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$
Where $$\alpha$$ and $$\beta$$ are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.
Answer
10
Explanation
$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$
$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$
$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$
$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$
$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$
At t = 0, $$\overrightarrow L = \overrightarrow 0 $$
$$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$$
t $$-$$ 2 (t $$-$$ 5) = 0
t = 10 sec
$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$
$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$
$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$
$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$
At t = 0, $$\overrightarrow L = \overrightarrow 0 $$
$$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$$
t $$-$$ 2 (t $$-$$ 5) = 0
t = 10 sec
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