A spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. If the top has a radius of 0.10 m, what is the centripetal acceleration of the edge of the top? 0.016 m/s2 0.73 m/s2 4.0 m/s2 6.3 m/s2

Question

Graham Woodward

Physics

3 May, 21:43

A spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. If the top has a radius of 0.10 m, what is the centripetal acceleration of the edge of the top? 0.016 m/s2 0.73 m/s2 4.0 m/s2 6.3 m/s2

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Dulce Flowers · Answer 1

Centripetal acceleration is defined mathematically as a = v^2 / r, where v is the velocity of the object and r is the radius of rotation. Centripetal acceleration is a vector and is always directed along the direction of the radius towards the centre of rotation. If the top has a radius of 0.10m and the speed at the edge of the spinning top is 0.62 m/s, then the answer in this case is 0.63^2 / 0.1 = 4.0 m/s^2.