Answer:
The angular momentum of a particle of mass m moving with velocity \[\vec{\upsilon }\] along a circular path of radius r is given by \[\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times m\vec{\upsilon }\] When the speed \[\upsilon \] decreases, the magnitude of angular momentum decreases. But the direction of angular momentum remains unchanged.
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