Question 455: Centre of Mass and Rotational Motion

Question

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.

Accepted Answer

$$\overrightarrow r  = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$
$$\overrightarrow v  = 20\alpha t\widehat i + 5\beta \widehat j$$
$$\overrightarrow L  = m(\overrightarrow r  	imes \overrightarrow v )$$
$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] 	imes [20\alpha t\widehat i + 5\beta \widehat j]$$
$$\overrightarrow L  = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$
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

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type