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NEET PHYSICSMedium

What is the value of linear velocity if ω⃗=3i^−4j^+k^\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k} and r⃗=5i^−6j^+6k^\vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}:

A

6i^+2j^−3k^6\hat{i} + 2\hat{j} - 3\hat{k}

B

−18i^−13j^+2k^-18\hat{i} - 13\hat{j} + 2\hat{k}

C

4i^−13j^+6k^4\hat{i} - 13\hat{j} + 6\hat{k}

D

6i^−2j^+8k^6\hat{i} - 2\hat{j} + 8\hat{k}

Step-by-Step Solution

  1. Formula: The relationship between linear velocity (v\mathbf{v}), angular velocity (ω\boldsymbol{\omega}), and the position vector (r\mathbf{r}) is given by the vector product (cross product): v=ω×r\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r} . Note: The order is important; r×ω\mathbf{r} \times \boldsymbol{\omega} would give the opposite sign.
  2. Calculation: Perform the cross product using the determinant method: v=∣i^j^k^3−415−66∣\mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\3 & -4 & 1 \\5 & -6 & 6 \end{vmatrix} v=i^((−4)(6)−(1)(−6))−j^((3)(6)−(1)(5))+k^((3)(−6)−(−4)(5))\mathbf{v} = \hat{i}((-4)(6) - (1)(-6)) - \hat{j}((3)(6) - (1)(5)) + \hat{k}((3)(-6) - (-4)(5)) v=i^(−24+6)−j^(18−5)+k^(−18+20)\mathbf{v} = \hat{i}(-24 + 6) - \hat{j}(18 - 5) + \hat{k}(-18 + 20) v=−18i^−13j^+2k^\mathbf{v} = -18\hat{i} - 13\hat{j} + 2\hat{k} .
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