At time t = 0,a 2, kg particle has position v | vedclass

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(A) The torque $\vec{	au}$ acting on a particle is given by $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes \frac{d\vec{p}}{dt} = \frac{d}{dt}(

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$32$

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(A) The torque $\vec{	au}$ acting on a particle is given by $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes \frac{d\vec{p}}{dt} = \frac{d}{dt}(\vec{r} 	imes \vec{p}) = \frac{d\vec{L}}{dt}$.Given mass $m = 2\, kg$,position $\vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v} dt = (4\hat{i} - 2\hat{j}) + \int_0^t (2t^2\hat{i}) dt = (4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}$.Velocity $\vec{v}(t) = 2t^2\hat{i}$.Angular momentum $\vec{L} = \vec{r} 	imes m\vec{v} = m [((4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}) 	imes (2t^2\hat{i})] = 2 [0 - 2\hat{j} 	imes 2t^2\hat{i}] = 2 [4t^2\hat{k}] = 8t^2\hat{k}$.Torque $\vec{	au} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(8t^2\hat{k}) = 16t\hat{k}$.At $t = 2\, s$,$\vec{	au} = 16(2)\hat{k} = 32\hat{k}\, N\cdot m$.

At time $t = 0$,a $2\, kg$ particle has position vector $\vec{r} = (4\hat{i} - 2\hat{j})\, m$ relative to the origin. Its velocity is given by $\vec{v} = 2t^2\hat{i}\, m/s$. The torque acting on the particle about the origin at $t = 2\, s$ is ........ $\hat{k}\, N\cdot m$.

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