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(B) The dot product of acceleration $\vec{a}$ and momentum $\vec{P}$ is given by: $\vec{a} \cdot \vec{P} = (4)(8) + (3)(-6) = 32 - 18 = 14 \, J/s$. Si

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Accelerated circular motion

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(B) The dot product of acceleration $\vec{a}$ and momentum $\vec{P}$ is given by: $\vec{a} \cdot \vec{P} = (4)(8) + (3)(-6) = 32 - 18 = 14 \, J/s$. Since the dot product $\vec{a} \cdot \vec{P} > 0$,the angle $	heta$ between the acceleration vector and the momentum vector is acute (less than $90^o$). The acceleration vector can be resolved into two components: one perpendicular to the momentum (centripetal acceleration) and one parallel to the momentum (tangential acceleration). Because the component of acceleration parallel to the momentum is positive,the magnitude of the velocity (speed) is increasing. Therefore,the motion is accelerated circular motion.

$A$ particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are $\vec{a} = (4\hat{i} + 3\hat{j}) \, m/s^2$ and $\vec{P} = (8\hat{i} - 6\hat{j}) \, kg \cdot m/s$. The motion of the particle is:

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