1. Jan 4, 2009

### KFC

I am quite confusing about the total kinetic energy of an rigid object swing in a vertical plane. Imagine a bar hanging to the ceiling with a massless but stretchable cord. The bar has mass M. Because the cord is stretchable (not very hard but like a spring with very small spring constant), the center of the bar $$(x_c, y_c)$$ changed.

Well if the cord is NOT stretchable, since the bar is swinging, the kinetic energy is $$K = \frac{I}{2}\omega^2$$. In this case, the center of the bar is also changing from place to place, but we don't account $$\frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)$$ into the total kinetic energy. But for the stretchable cord, the center of the bar is also changing but in other way, in this case, what will the total kinetic energy look like? Will it be

$$K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)$$

But someone tell me it is not necessary to include the term about $$x_c$$, only the following is enough

$$K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2$$

This is very confusing! For rotation, when should we include the translation kinetic energy? and when could we use the rotation kinetic energy only for the whole system?

By the way, if we use a mass instead, what would be different?

2. Jan 5, 2009

### tiny-tim

Hi KFC!

You're getting confused about whether I is measured relative to an axis through the ceiling or through the centre of mass.

From the PF Library:
This is the same as your formula if you put x'2 + y'2 = v2 = w2d2.
Sorry, don't understand that.

3. Jan 5, 2009

### Lucien1011

Sometimes I think it is useful to think of mv2/2 for each point first.