Physics help???

Question

a bead slides without friction around a loop-the-loop. the bead is released from a height of 24.7 m fromt he bottom of the loop-the-loop which has a radius 9m. the acceleration of gravity is 9.8m/s/s.

what is its speed at the top of the loop-the-loop? answer in units of m/s.

Answer 1

this answer does not take into account gravity when it reaches the top of the loop, it is effected by gravity at some point

use potential energy to kinetic energy to find velocity into the loop:
mgh = (1/2)mv^2
mass falls out to get
v = sqrt(2*9.8*24.7)
v = 22 m/s
this answer is the velocity into the loop, v=22m/s

once in the loop it would be in radians/second
v = radius*angular velocity
v = rw
w = 22/9 = 2.44 rad/s
this is as close as i could get

Answer 2

AA was on a good approach but missed one trick. For the potential energy you can use the elevation of the starting point over the top of the loop. Yes, the velocity will be highest at the bottom of the loop, but then some of that kinetic energy is converted back to potential energy. So use
h = 24.7 m - 2 times the radius of the loop.
h = 6.7 m

Then follow AA's steps to get velocity and you're done.