A marble is going around a frictionless vertical circular track of radius 1 mete

Question

A marble is going around a frictionless vertical circular track of radius 1 meter as shown. What is the minimum speed that the ball must have at the top of the track is it is not to fall off when it passes over the top of the track? (express your answer in m/s)

Hint: Anywhere on the track, the marble needs to have a net centripetal force appropriate for its speed and the radius of the track. At the top of the track, the net centripetal force must be at least as strong as the gravitational force pulling down.

Explanation / Answer

here,

radius , r = 1 m

let the speed at the top be

for the marble not to fall

m * v^2 /r = m * g

v = sqrt(r * g)

v = sqrt( 1 * 9.81) m/s

v = 3.13 m/s

the speed of marble at the top is 3.13 m/s

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