Regarding total rotational kinetic energy

Question

The rotational kinetic energy for a body that is rolling is is $\boldsymbol{\frac{1}{2}Iω^2}$ (where $I$ is moment of inertia about its centre of mass) and the translational kinetic energy is $\boldsymbol{\frac{1}{2}mv^2}$ for a rolling body..where $v$ is speed of its centre of mass for an inertial observer

If we add both of these for the body shown in the figure then we should get its total kinetic energy at a particular instant.

$\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$

this should be a generally applicable formula because I have seen the derivation and it seems to be applicable for any rolling body...

But it is yielding the wrong answer in this case..can anyone please tell me why?

P.S : Sorry for bad circle in the top view.

Answer 1

Here, though it is a rigid body, you cannot use $KE_{TOT} = \frac{1}{2}M{v^2}_{cm}+\frac{1}{2}I\omega ^2$ because the particles closer to the larger axis (Radius $R$) are moving slower than those far away. So we must find KE_TOT as :

$KE_{TOT} = \frac{1}{2}I_o{\omega_o}^2 + \frac{1}{2}I_p{\omega_p}^2$........(1)

The moment of inertia of sphere about $O$ is $\frac{2}{5}Mr^2 + MR^2$ and $\omega_o$ is $\frac{V}{R}$

Moment of inertia about $P$ is $\frac{2}{5}Mr^2$ and $\omega_p$ is $\frac{V}{r}$

substituting into (1)

$KE_{TOT}$

$= \frac{1}{2}\left(\frac{2}{5}Mr^2 + MR^2 \right) {\left(\frac{v}{R}\right)} ^2 + \frac{1}{2}\left(\frac{2}{5}Mr^2\right ) {\left(\frac{v}{r}\right )}^2$

$=\frac{7}{10}Mv^2 + \frac{1}{5}\frac{r^2}{R^2}v^2$

which is the correct result.

Answer 2

Kinetic energy equal to $\frac{1}{2}mV^2$, in general is not true for bodies having finite dimensions in non rectilinear motion. Use parallel axis theorem to find $K.E.$ for the revolution of the sphere. However if the sphere was point size then I think your point is valid.

Answer 3

Nice problem!

Note, that the angular velocity of the sphere is NOT $V/r$. There is a component of the angular velocity directed along the vertical axis.

Imagine another, simpler problem. It's almost the same, but the sphere is not rolling on a table. It is sliding along it.

It's kinetic energy would not be $V^2/R$. Because the sphere is actually rotating - you will see it if look at the sphere from above!

UPDATE.

Formula $E=mv^2 + I\omega^2/2$ (where $v$ is the velocity of center of mass) is correct. In this particular problem it is very easy to make a mistake calculating the angular velocity $\omega$ and so get incorrect final answer.

Looks like the axle of rotation of the sphere at any moment is $OP$ - the line which goes via $O$ and center of the sphere $P$. But this is not actually so!

In the frame of reference which does not rotate but is moving with the same velocity $\vec{v}$ as the center of mass, velocity of any point of the body is $\vec{v}(\vec{r}) = [\vec{w}*\vec{r}$] , where $\vec{r}$ is a vector from the center of mass to our point of the body. For all the points along the axle of rotation this velocity is zero.

In the original frame of reference all these points should have the same velocity (same as the velocity of the center of mass).

But clearly the velocities of different points of sphere located along the axle $OP$ are different - further from $O$, bigger the speed. So, $OP$ is not the axle of rotation of the sphere!

Well, if you get into the frame of reference which rotates around point $O$ with angular velocity $W=V/R$ the speed of every point along the $OP$ would be zero. This would be the axle of rotation of sphere, and in this frame of reference the angular velocity of the sphere would indeed be $V/r$. And to find the angular velocity in the original frame of reference you need to add up $\vec{w}$ and $\vec{W}$ - but remember that both of them are vectors and you should add them as vectors!