An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at x = + 2 m , its velocity is - 4 j ^ m s - 1 . The object’s velocity ( v

Question

An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at x = + 2 m , its velocity is - 4 j ^ m s - 1 . The object’s velocity ( v ) and acceleration ( a ) at x = – 2 m will be

TestHub · Accepted Answer

As the speed is constant, only centripetal acceleration will act(towards the centre of the circular path), $a_{c} = \frac{v^{2}}{r} = \frac{4^{2}}{2} = 8 m s^{- 2}$ towards the centre which at the given time will be towards right, hence ${\vec{a}}_{c} = 8 \hat{i} m s^{- 2}$ . Velocity of the particle at this instant will in upward direction and hence $\vec{v} = 4 \hat{j} m s^{- 1}$ .

Physics - Circular Motion Question with Solution | TestHub

Options:

Doubts & Discussion