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Rotational Motion Part 3

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Section: Physics

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Question : 42 of 62

Marks: +1, -0

The position vectors of two 1 kg particles,

(A)

(B)

\overset{\rightarrow_{A}}{r}= (\alpha_1 t^{2} \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k})\ \mathrm{m}

\;\;\overset{\rightarrow_{B}}{r}= (\beta_1 t \hat{i} + \beta_2 t^{2} \hat{j} + \beta_3 t \hat{k})\ \mathrm{m}

, respectively;

\alpha_1 = 1\ \mathrm{m} / \mathrm{s}^{2},\ \alpha_2 = 3\ \mathrm{m} / \mathrm{s}^{2},\ \alpha_3 = 2\ \mathrm{m} / \mathrm{s},\ \beta_1 = 2\ \mathrm{m} / \mathrm{s}.

\beta_2 = -1\ \mathrm{m} / \mathrm{s}^{2},\ \beta_3 = 4p\ \mathrm{m} / \mathrm{s}

t

n

p

are constants. At

t = 1\ \mathrm{s},\ \overset{\rightarrow_{A}}{|V|} = \overset{\rightarrow_{B}}{|V|}

\overset{\rightarrow_{A}}{V}

\overset{\rightarrow_{B}}{V}

of the particles are orthogonal to each other. At

t = 1\ \mathrm{s}

, the magnitude of angular momentum of particle

(A)

with respect to the position of particle (B) is

\sqrt{L}\ \mathrm{kgm}^{2}\ \mathrm{s}^{-1}

L

[22 Jan 2025 Shift 1]

Your Answer:

Solution: 👈: Video Solution

\;\;\text{ At }\; t = 1

r_{AB} = -1 \hat{i} + (3n+1) \hat{j} + (2-4p) \hat{k}

\;\;\text{ At }\; t = 1

v_A = 2 \hat{i} + 3n \hat{j} + 2 \hat{k}

v_B = 2 \hat{i} - 2 \hat{j} + 4p \hat{k}

\overset{\rightarrow_{A}}{v} \cdot \overset{\rightarrow_{B}}{v} = 0,\;\; 4 - 6n + 8p = 0

|v_A| = |v_B|\;\; (3n)^2 + 4 = 4 + 16p^2

3n = -4p

4 + 16p = 0

p = -\frac{1}{4},\; n = \frac{1}{3}

r_{AB} = -\hat{i} + 2 \hat{j} + 3 \hat{k}

v_A = 2 \hat{i} + \hat{j} + 2 \hat{k}

\therefore\ L = m \overset{\rightarrow}{\left| r_{AB} \times \overset{\rightarrow_{A}}{v} \right|} = 90

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