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A body is free to rotate about an axis passing through the y-axis. A force of $\vec{F} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) N$ is acting on the body the position vector of whose point of application is $\vec{r} = (2 \hat{i} - 3 \hat{j}) m$ . The moment of inertia of body about y-axis is 10 kg m². Then the angular acceleration of the body should be equal to:

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$(- 1.8 \hat{i} - 1.2 \hat{j} + 1.3 \hat{k}) rad s^{- 2}$

$- 1.8 \hat{i} {rads}^{- 1}$

$- 1.2 \hat{j} {rads}^{- 2}$

$1.3 \hat{k} {rads}^{- 2}$

answer is C.

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Detailed Solution

A body is free Torque about origin experienced by body is

$\begin{array}{l} \vec{τ} = \vec{r} \times \vec{F} = (2 \hat{i} - 3 \hat{j}) \times (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) N - m \\ = - 18 \hat{i} - 12 \hat{j} + 13 \hat{k} \end{array}$

But as the body is allowed to rotated only along y-axis, the y-component of torque only causes the angular acceleration of the body. So

$α = \frac{τ_{y}}{I} = \frac{- 12}{10} \hat{j} = - 1 .2 \hat{j} rad / s^{2}$

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