Work and energy - easy question

[7tongc5]

a mass "m" slides without friction along a looped apparatus. If the object is to remain on the track, even at the top of the loop (whose radius is "r") from what minimum height "h" must it be released?

 

[arildno]

Remember that at the top of the loop, gravity must provide the centripetal acceleration of the object.

 

[7tongc5]

right. so

mg = (mv^2)r, masses cancel, solve for v^2

v^2 = 9.8r

and E = 1/2mv^2 + mgh
mgh = 1/2m (9.8r) + m(9.8)(2r)
9.8h = 1/2 (9.8r) + (9.8)(2r)
h = 1.55 --> = height of loop

am i tackling this right so far?

 

[arildno]

right. so
mg = (mv^2)r, masses cancel, solve for v^2
v^2 = 9.8r
and E = 1/2mv^2 + mgh
mgh = 1/2m (9.8r) + m(9.8)(2r)
9.8h = 1/2 (9.8r) + (9.8)(2r)

Right, when "h" is measured from the level of the bottom of the loop.

h = 1.55 --> = height of loop
am i tackling this right so far?

just divide with g, and get h=2.5r.