1/2 m v^2 = m g h
v = √(2 g h) = √(2 * 9.8 * .45) m/s
A ball is moving to and fro about the lowest point of a smooth hemispherical bowl. If it is able to rise up to a height of 45cm on either side.it's speed at the lowest point must be?
3 answers
humm, is it sliding or rolling?
If rolling then some of that potential energy goes into kinetic at the bottom
Ke = (1/2) m v^2 + (1/2) I omega^2
I = (2/5)mr^2 for solid ball
so
Ke = (1/2)mv^2 + (1/2)(2/5)mr^2 omega^2
but v = r omega
so
Ke = (7/10)mv^2
so
9.8*.45 = .7 v^2
by the way if a thin walled sphere, then I = (2/3)mr^2
If rolling then some of that potential energy goes into kinetic at the bottom
Ke = (1/2) m v^2 + (1/2) I omega^2
I = (2/5)mr^2 for solid ball
so
Ke = (1/2)mv^2 + (1/2)(2/5)mr^2 omega^2
but v = r omega
so
Ke = (7/10)mv^2
so
9.8*.45 = .7 v^2
by the way if a thin walled sphere, then I = (2/3)mr^2
gain in potential energy=loss in kinetic energy
so 1/2mv^2=mgh
v^2=gh
v=underroot of 2gh
=underroot of 2*10*.2
=2m/s
so 1/2mv^2=mgh
v^2=gh
v=underroot of 2gh
=underroot of 2*10*.2
=2m/s