A force overrightarrow{F} = alpha hat{i} + 3h | vedclass

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(B) For the conservation of angular momentum about the origin,the net torque $\overrightarrow{	au}$ acting on the particle must be zero.By definition

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$-1$

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(B) For the conservation of angular momentum about the origin,the net torque $\overrightarrow{	au}$ acting on the particle must be zero.By definition,$\overrightarrow{	au} = \overrightarrow{R} 	imes \overrightarrow{F}$.Given $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ and $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$.Calculating the cross product:$\overrightarrow{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -6 & -12 \ \alpha & 3 & 6 \end{vmatrix}$$\overrightarrow{	au} = \hat{i}(-36 - (-36)) - \hat{j}(12 - (-12\alpha)) + \hat{k}(6 - (-6\alpha))$$\overrightarrow{	au} = \hat{i}(0) - \hat{j}(12 + 12\alpha) + \hat{k}(6 + 6\alpha)$For $\overrightarrow{	au} = 0$,both the $\hat{j}$ and $\hat{k}$ components must be zero:$12 + 12\alpha = 0 \implies \alpha = -1$$6 + 6\alpha = 0 \implies \alpha = -1$Thus,the value of $\alpha$ is $-1$.

$A$ force $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$ is acting at a point $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is

Similar Questions

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