The time dependence of the position of a part | vedclass

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(A) Given: $\vec r(t) = 2t \hat i - 3t^2 \hat j$ and $m = 2 \ kg$.Velocity $\vec v = \frac{d\vec r}{dt} = 2 \hat i - 6t \hat j$.At $t = 2 \ s$,positio

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$-48 \hat k$

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(A) Given: $\vec r(t) = 2t \hat i - 3t^2 \hat j$ and $m = 2 \ kg$.Velocity $\vec v = \frac{d\vec r}{dt} = 2 \hat i - 6t \hat j$.At $t = 2 \ s$,position $\vec r = 2(2) \hat i - 3(2)^2 \hat j = 4 \hat i - 12 \hat j$.At $t = 2 \ s$,velocity $\vec v = 2 \hat i - 6(2) \hat j = 2 \hat i - 12 \hat j$.Momentum $\vec p = m \vec v = 2(2 \hat i - 12 \hat j) = 4 \hat i - 24 \hat j$.Angular momentum $\vec L = \vec r 	imes \vec p = (4 \hat i - 12 \hat j) 	imes (4 \hat i - 24 \hat j)$.Using the cross product: $\vec L = [4(-24) - (-12)(4)] \hat k = (-96 + 48) \hat k = -48 \hat k$.

The time dependence of the position of a particle of mass $m = 2 \ kg$ is given by $\vec r(t) = 2t \hat i - 3t^2 \hat j$. Its angular momentum with respect to the origin at time $t = 2 \ s$ is:

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