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.
Answer (integer)
10
Solution
$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$<br><br>$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$<br><br>$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$<br><br>$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$<br><br>$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$<br><br>At t = 0, $\overrightarrow L = \overrightarrow 0$<br><br>$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$<br><br>t $-$ 2 (t $-$ 5) = 0<br><br>t = 10 sec
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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