00:02
Assume we have a pendulum consisting of a chord of length l and a mass m attached to it.
00:11
If this pendulum is allowed to swing and is then interrupted at its lowest point by a peg, the pendulum will transcribe a circle in its motion.
00:25
How do we know what the minimum possible speed at the bottom of its motion, or the pendulum would be to be able to put it.
00:35
Pass through the entire circle.
00:38
Well, let's look at what the conservation of energy tells us.
00:44
Firstly, at the bottom of the circle, this is the initial point of reference.
00:52
We know u, which is the potential energy, mgh, is zero.
01:00
So the bottom of the circle is taken to be at a height zero.
01:07
And the kinetic energy, k, is equal to a half m v i square and this is what we are trying to find v i the velocity as the pendulum begins to make its circular trajectory at the top for our final reference position we know u is equal to m g h and h in this case is just l minus.
01:50
So at this point at the top of the circle, what is the kinetic energy at this point? k is equal to a half mv square the final velocity.
02:11
Well we know that this is centripetal motion.
02:18
So the kinetic energy is m g times r.
02:25
This is due to the fact that this interpidate acceleration g is just v squared divided by r and r being the radius of the circle is just l minus d so that's a half m g into l minus d now the conservation of energy tells us that the energy before has to be the mechanical energy after so a half m v i squared is equal to two mg into l minus d plus a half ng into l minus d.
03:19
So the sum of the kinetic energy and potential energy before, or at the bottom of the circle, has to give us the sum of the potential and kinetic energies at the top of the circle.
03:32
So if we do a bit of rearranging, we find vi the square root of 5 times gravitation acceleration g into the radius of the circle l minus d...