A bead is released from a height of 25.2 m, travels down (25.2 m) & slides without friction around a loop-the-loop, which has a radius of 9 m. What is the speed at the highest point on the loop the loop (in m/s)? Acceleration of gravity is m/s^2.
Isn't the mass of the bead needed to solve? If not, how?
2006-10-12 11:02:14 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
This is a simple conservation of energy problem. All that matters is the change in height. Equating PE and KE you get
mgh = mv^2/2. Note that the mass cancels and you get
v=sqrt(2gh)
From the text h=25.2-2*9 = 7.2 m, so
v = 11.9 m/s
2006-10-12 11:14:53 · answer #1 · answered by d/dx+d/dy+d/dz 6 · 0⤊ 0⤋
you do not need mass because you assume the bead is in freefall due to the lack of friction. You can then use UVATS equations to find the speed of the bead at certain heights.
2006-10-12 11:06:10 · answer #2 · answered by Stuart T 3 · 0⤊ 0⤋