A little question

[AlanR]

hi guys
somehow i dont quite get this, maybe someone can clarify it for me...

the apparatus
a simple pendulum, mass = 1Kg, length = 1 metre.
mass is at start position, ready to swing from height same as pivot point, 1 metre .
OK, so the potential energy = MGH = 1 * 9.8 * 1 = 9.8 J
Now, pendulum is released, accelerates and at lowest point (zero PE) has energy = 0.5mvv. = 9.8 J , same as PE was at start point, OK simple.

However, doesn't the mass now also have rotational kinetic energy (as angular momentum) about its own axis? ie, if the string were cut at this point wouldn't the mass continue to rotate about its own axis as it flew off at the tangent? (but fall from gravity as well of course).

So, where does the energy 'absorbed' by the mass into its own angular momentum come from with regard to the conventional PE > KE > PE equations?

I wish there were some magic here in that PE + KE = constant, but the angular rotantional energy appears from "nowhere". Dream on, huh?

Anyway, does anyone have the actual balanced equations that take this into account? Or am i talking rubbish?

thanks guys

Alan.

[Jonathan]

The equation mgh=.5mvv assumes the mass is essentially a point. To be more accurate, especially when the mass is physcially large, you'd have to take the rotational energy into account too.

[Jonathan]

I did some math, and if you have a uniform circle of mass m and radius r acting as a pendulum bob for a pendulum of length R, and the bob is started at pivot height, then
w_1=sqrt(2gR/(.5rr+RR))
w_2=2w_1
v=sqrt(2(gR-rr(w_1)(w_1))))
where w_1 is "omega (subscript) one" and is the angular velocity of the pendulum at the nadir, w_2 is the angular velocity of the bob if released at that point, and v is its linear velocity if released at that point.
EDIT
The circle is mounted to be in the same plane as that of rotation.