Work and energy - easy question

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To ensure a mass "m" remains on a frictionless track at the top of a loop with radius "r," it must be released from a minimum height "h." The centripetal force required at the top is provided by gravity, leading to the equation mg = (mv^2)/r, which simplifies to v^2 = 9.8r. The total energy equation, E = 1/2mv^2 + mgh, is used to relate height and speed, resulting in h = 1.55 when calculated from the loop's bottom. However, further simplification suggests that the correct minimum height should be h = 2.5r. The calculations confirm the relationship between height and radius for maintaining motion through the loop.
7tongc5
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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?
 
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Remember that at the top of the loop, gravity must provide the centripetal acceleration of the object.
 
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?
 
7tongc5 said:
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?
:confused:

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