The figure is below. Let 1 be the highest position on the loop, 2 be the lower position on the loop and 3 the release point.

The condition for being even at the top of the track is

`F_cf = G`

`(m*v_1^2)/R =m*g rArr v_1^2 =gR`

Because there is no...

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The figure is below. Let 1 be the highest position on the loop, 2 be the lower position on the loop and 3 the release point.

The condition for being even at the top of the track is

`F_cf = G`

`(m*v_1^2)/R =m*g rArr v_1^2 =gR`

Because there is no friction the total energy is the same at the top and at the bottom of the loop.

`(mv_1^2)/2 +mg*(2R) = (m*v_2^2)/2`

`v_2^2 =v_1^2+4gR`

`v_2^2 =gR +4gR =5gR`

Also on the part of the falling ramp from point 3 to 1 there is no friction and the total energy is the same.

`(mv_2^2)/2 =mgh rArr h=v_2^2/(2g) =(5gR)/(2g) =(5/2)*R`

**Answer: the mass has to be released from a height of `(5/2)R` to be even at the top of the loop**.