A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of the loop in order to remain in contact with the loop

3.834 m/s. In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is F = mv^2/r and the equation for gravitational attraction is F = ma Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so a = v^2/r Substitute the known values (radius is diameter/2) and solve for v 9.8 m/s^2 < = v^2/1.5 m 14.7 m^2/s^2 < = v^2 3.834057903 m/s < = v So the minimum velocity needed is 3.834 m/s.