A particle moves in the xy-plane under the ac | vedclass

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(A) Given the linear momentum vector is $\vec{p} = p_x \hat{i} + p_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.The force $\vec{F}$ is the rate

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$90$

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(A) Given the linear momentum vector is $\vec{p} = p_x \hat{i} + p_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.The force $\vec{F}$ is the rate of change of linear momentum,so $\vec{F} = \frac{d\vec{p}}{dt}$.Calculating the derivative: $\vec{F} = \frac{d}{dt}(2\cos t) \hat{i} + \frac{d}{dt}(2\sin t) \hat{j} = (-2\sin t) \hat{i} + (2\cos t) \hat{j}$.To find the angle between $\vec{F}$ and $\vec{p}$,we calculate their dot product: $\vec{F} \cdot \vec{p} = [(-2\sin t)(2\cos t)] + [(2\cos t)(2\sin t)]$.$\vec{F} \cdot \vec{p} = -4\sin t \cos t + 4\sin t \cos t = 0$.Since the dot product $\vec{F} \cdot \vec{p} = |\vec{F}| |\vec{p}| \cos 	heta = 0$,and neither vector is a null vector,we must have $\cos 	heta = 0$,which implies $	heta = 90^o$.

$A$ particle moves in the $xy$-plane under the action of a force $\vec{F}$ such that the components of its linear momentum $\vec{p}$ at any time $t$ are $p_x = 2\cos t$ and $p_y = 2\sin t$. The angle between $\vec{F}$ and $\vec{p}$ at time $t$ is ........... $^o$.

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