A particle having mass 4 kg is moving with v | vedclass

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Question

(A) The position vector of the particle is $\vec{r} = (1 \hat{i} + 1 \hat{j} + 0 \hat{k}) \ m$.The velocity vector is $\vec{v} = (4 \hat{i} + 2 \hat{j

Vedclass · Accepted Answer

$-8 \hat{k} \ kg \cdot m^2/s$

Vedclass · Answer

(A) The position vector of the particle is $\vec{r} = (1 \hat{i} + 1 \hat{j} + 0 \hat{k}) \ m$.The velocity vector is $\vec{v} = (4 \hat{i} + 2 \hat{j}) \ m/s$ and mass $m = 4 \ kg$.The angular momentum $\vec{L}$ is given by the cross product $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m \vec{v})$.Substituting the values: $\vec{L} = (\hat{i} + \hat{j}) 	imes [4 	imes (4 \hat{i} + 2 \hat{j})]$.$\vec{L} = (\hat{i} + \hat{j}) 	imes (16 \hat{i} + 8 \hat{j})$.Calculating the cross product: $\vec{L} = 16(\hat{i} 	imes \hat{i}) + 8(\hat{i} 	imes \hat{j}) + 16(\hat{j} 	imes \hat{i}) + 8(\hat{j} 	imes \hat{j})$.Since $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{j} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{j} 	imes \hat{i} = -\hat{k}$.$\vec{L} = 0 + 8(\hat{k}) + 16(-\hat{k}) + 0 = 8 \hat{k} - 16 \hat{k} = -8 \hat{k} \ kg \cdot m^2/s$.

$A$ particle having mass $4 \ kg$ is moving with velocity $(4 \hat{i} + 2 \hat{j}) \ m/s$. Find the angular momentum of the particle about the origin when it is at position $(1, 1, 0) \ m$.

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