Total energy of system along axis

[alterecho]

Its stated that total energy of an isolated system is given by
E = $\frac{1}{2}$m$v^{2}$ + $\frac{1}{2}$I$ω^{2}$ + mgh.

Whats the correct form when the velocity is along 2 axis, that is, along x and y-axis like v = (2, 1)?
Should the resultant be taken or just add the kinetic energies along each axis?

I've been searching awhile but can't find an accurate answer.

thanks in advance.

 

[Doc Al]

Whats the correct form when the velocity is along 2 axis, that is, along x and y-axis like v = (2, 1)?
Should the resultant be taken or just add the kinetic energies along each axis?

The translational KE would be:
$$KE = 1/2mv^2 = 1/2m(v_x^2 + v_y^2)$$

 

[alterecho]

Thanks. I have a further related question. Now suppose i want to extract the final velocity given the initial total energy, the final rotational energy (assuming the potential energies are 0 in this case) using the law of conservation of energy given by,

$\frac{1}{2}$m$u^{2}$ + $\frac{1}{2}$I$ω^{2}$ = $\frac{1}{2}$m$v^{2}$ + $\frac{1}{2}$I$W^{2}$

or

m$u^{2}$ + I$ω^{2}$ = m$v^{2}$ + I$W^{2}$

since initial energy of the system is known, left side of the equation becomes $E_{i}$,
$E_{i}$ = m$v^{2}$ + I$W^{2}$

since m, I and $W_{2}$ are known, i get the equation for finding the velocity as,
v = $\sqrt{\frac{E_{i} - IW^{2}}{m}}$

But how do i break it down into x and y coordinates?

 

[Doc Al]

But how do i break it down into x and y coordinates?

There's no way to tell the final velocity direction from conservation of energy alone. You'll have to know the details of the constraint forces involved. (Energy is a scalar, not a vector.)

 

[alterecho]

Oh. But I've read that for resolution of collision, they take into consideration energy conservation. So how would they resolve it? Are we supposed to know the initial and final velocities and then calculate the final angular velocity from that?