The position of a particle is given by r- - i-2j-k and momentum P- - 3i-4j-2k- The direction of angular momentum is perpendicular to

L⃗ = r⃗×p⃗ = i amp; j amp; k 1 amp; 2 amp; 1 3 amp; 4 amp; 2 = 0i j 2k = j 2k and the X axis is given by i + 0j + 0k ...

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Angular Analogue of Linear Momentum

The position ...

Question

The position of a particle is given by $\to r = (ˆ i + 2 ˆ j - ˆ k)$ and momentum $\to P = (3 ˆ i + 4 ˆ j - 2 ˆ k)$ . The direction of angular momentum is perpendicular to

X-axis

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Y-axis

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Z-axis

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Line at equal angles to all the three axes

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Solution

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2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}

3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k} and - \hat{i} + 4 \hat{j} - 2 \hat{k}

\to r = (ˆ i + 2 ˆ j - ˆ k)

\to P = (3 ˆ i + 4 ˆ j - 2 ˆ k)

\to r = (^i + 2^j -^k) m

\to p = (3^i + 4^j - 2^k) kg-m/s

(0, 0, 0)

\to p

\to r

\to L

\to r

\to p

\to L

m = 6 k g

\to r = [(3 t^{2} - 6 t)^i + (- 4 t^{3})^j] m

(\begin{matrix} \to F = m \to a \end{matrix})

(\begin{matrix} \to τ = \to r \times \to F \end{matrix})

(\begin{matrix} \to p = m \to v \end{matrix})

(\begin{matrix} \to L = \to r \times \to p \end{matrix})

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