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.

enter image description here

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

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.

  • $\begingroup$ You reasoning seems correct and it should give correct. Can you show your work in brief? $\endgroup$
    – sslucifer
    Commented May 17, 2020 at 19:21
  • $\begingroup$ @sslucifer-My answer is coming out to be .7/ 10 mv^ 2 .I simply added the rotational kinetic energy for a rolling sphere and its translational kinetic energy..but the correct answer Is bigger in magnitude I.e 7/10 mv^2 ( 1 + 2r^2/7R^2 )..according to me, r and R should not even be ending up as variables in the final answer..well I can't seem to wrap my head around this one.. $\endgroup$ Commented May 17, 2020 at 19:49

3 Answers 3


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)

*refer diagram*

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)


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

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

which is the correct result.

  • $\begingroup$ yes..I think I now understand what you are saying...thank you for your answer : )I need to still keep thinking about it..but it now makes sense because you said particles closer to the central axis are moving slower and the derivation that I read does not fit with that fact....but hey..can you read my comment below regarding the 'magic track' situation and explain if the kinetic energy will increase once the sphere is moving on the track ..which would be very counter intuitive..I personally think now that the velocity of centre of mass will change then..to keep K.E constant. $\endgroup$ Commented May 18, 2020 at 6:58
  • $\begingroup$ There is no magic out there. we may see rolling as two motions acting together - a rotation and a translation Initially, when the sphere was rolling, every particle had same speed(keeping aside speed due to rotation) But as you connect a rod, now only the central point has the same speed. Points farther away now has a higher velocity and points closer are moving slower. $\endgroup$ Commented May 18, 2020 at 7:26
  • $\begingroup$ Now you may say- some gained velocity some lost velocity. So shouldn't tge total energy be same? But no... we know velocities increase propotional to distance from axis. if $R$ is centre of mass, $v(R+\Delta R) =(R+Delta R)\omega$ and $v(R-\Delta R) =(R-Delta R)\omega$. Find kinetic energies of these particles. It would be proportional to $(R+\Delta R)^2$ and $(R-Delta R)^2$. Find their sum. You would see that the net kinetic energy has increased from $R^2$ to $R^2 + (\Delta R)^2$ $\endgroup$ Commented May 18, 2020 at 7:36
  • $\begingroup$ So from where the extra energy? we have to give that to change the state of motion from completely straight to curved $\endgroup$ Commented May 18, 2020 at 7:39
  • $\begingroup$ @ rishab navneet..okay..yes your explanation is quite visual and intuitive.thank you : ) I think I've understood where I was going wrong after you mentioned the R +(dR)^2 proportionality to the d(K.E)..but if I may ask one more question..which force do you think does the additional work once the sphere starts rolling on the curved track.because there needs to be some force doing work to cause the extra kinetic energy to come about.. the normal forces between the track and the sphere's body would do the work initially right. if I may guess .? $\endgroup$ Commented May 18, 2020 at 9:25

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.

  • $\begingroup$ okay..you have a point..now I have to think about this along with Rishabh navneet's answer.thank you : ) $\endgroup$ Commented May 18, 2020 at 6:50
  • $\begingroup$ @ lesnik..hey just read your update..hmm your answer is another way of thinking about the problem..different from rishab navaneet's.it also seems right from a first principles viewpoint i.e from the perspective of the derivation for the kinetic energy equation.i need some time to think about your and rishab's answers..but I have an intuition that both are equivalent. Thanks a lot : ) $\endgroup$ Commented May 18, 2020 at 9:33
  • $\begingroup$ @ lesnik and @ rishab navaneet : both of your answers are very cool though.thank you : ) $\endgroup$ Commented May 18, 2020 at 9:39

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!


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!

  • $\begingroup$ @ lesnik: yes but the derivation for kinetic energy of a rolling body says that doesn't matter whether the body is rolling about some x,y,y point in a circular track just take the velocity of the center of mass how so ever you get it and if the body is pure rolling( which it is here) then we automatically get the angular velocity too and voila we can get the total kinetic energy..also in your example if sphere is simply sliding on a table..its kinetic energy will only be 1/ 2 mv^2 where v will be its tangential speed at any instant.. $\endgroup$ Commented May 17, 2020 at 20:57
  • $\begingroup$ @ lesnik: .if you think I need a better understanding..it would be helpful if you could provide maybe a link..in case I'm missing something bigtime..or maybe you could elaborate your previous point. $\endgroup$ Commented May 17, 2020 at 20:57
  • $\begingroup$ @ lesnik: Imagine this..let us say the sphere is rolling on the table freely..its kinetic energy during this rolling without being connected to anything is 7/10 mv^2...now suddenly a circular rail track appears by 'magic' and there is a small entry point on the track.like on a freeway. the sphere enters the track and starts going around in a circular motion around the centre point of the track..assuming no frictional losses..shouldn't its kinetic energy still be the same as before..the normal forces on the track's sides won't do any work clearly.. $\endgroup$ Commented May 17, 2020 at 21:04

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