I was solving a practice test problem and it was just a conservation of energy problem where a spherical ball is falls from a height h to the ground such that

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

what I can't quite picture in my head is how there is an energy contribution from the rotation at a fixed point in time. obviously there is a instantaneous rate of change of the center of mass

$$v_{cm} = \frac {d \vec {x_{cm}} }{dt}$$ for any fixed point in time in my head I can see this as contributing to energy

but the angular velocity can't be similarly treated like it is all at a point $$ \vec{\omega} = \frac{d\ \vec{\theta} } {dt}$$

here i understand this vector of angular rotation exists but I am confused as to why it adds to the energy of the system

I just somehow can't wrap my head about energy of rotation when there is already some energy due to the center of mass' translation... i'm not sure exactly why sorry if this is vague but I figured i'd ask if someone has some way to think about this. maybe in terms of DOF's?

  • $\begingroup$ obviously, if we were to consider say a spinning top with $v_{cm}$ = 0 there would obviously be some energy there... $\endgroup$ – Timtam Aug 10 '11 at 9:44

Try looking at this problem microscopically:

You can imagine the ball consisting of a number of smaller pieces of matter. The total kinetic energy of the entire ball is the sum of the kinetic energies of its pieces:


Now, if it were to just fall straight down, all these little pieces would have velocity vector $v_i=v$ and you could write:

$$T=\frac{1}{2}(\sum_im_i)v^2 = \frac{1}{2}Mv^2$$

where $M$ is the total mass.

But if it is rotating, the small pieces have an additional component in the velocity: they do not just move downwards, but have also to move around the center of rotation of the body.

So in addition to the downward falling component of the motion, the particles have:

$$T_{rot}=\frac{1}{2}\sum_im_iv_{\text{rot},i}^2 = \frac{1}{2}\sum_im_i(\omega r_i)^2 = \frac{1}{2}\omega^2\sum_im_ir_i^2 = \frac{1}{2}\omega^2 I$$

so the total energy is:

$$T_{tot}=\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\omega^2 I$$


The motion of a rigid body can be viewed as the sum of two parts: a translation of the center of mass plus rotation about the center of mass. From this, the velocity of any point $i$ of the body relative to the fixed system of coordinates is the sum of the translational velocity of the CM and the angular velocity of rotation around the CM: $$\mathbf{v_i} = \mathbf{V}_i + \boldsymbol{\omega} \times \mathbf{r}_i.$$ The total kinetic energy of the moving body is the sum of the kinetic energies of all discrete particles constituting the body: $$T = \sum_{i} \frac{1}{2}m_i \mathbf{v}_i^2.$$ Substituting $\mathbf{v}_i$ in the above equation (and doing some magic) gives $$T = \frac{1}{2} MV^2 + \frac{1}{2} \sum_{i} m_i ( \omega^2 r_i^2 - (\boldsymbol{\omega} \mathbf{r_i})^2 ).$$ The first term is the kinetic energy of the translational motion of body's CM, while the second is the kinetic energy of rotation (with angular velocity $\boldsymbol{\omega}$) about the axis passing through the CM. In the simplest case of rotation about a fixed axis of the body's symmetry the second term will look like you wrote: $\frac{1}{2}I \omega^2$.


Think of it as this,

If you are having a ball, which is rotating with say v and angular velocity $ \omega $, then consider that both of these velocities constitute to its total energy E.

$$ E = E{rot} + E{trans} $$

Now, consider the case when either of them is 0, then since both things are independent of each other, it makes sense to assume that other will be unaffected.

and again we know that, $$ E{rot} = 1/2 m v^{2} $$ $$ E{trans} = 1/2 I\omega^{2} $$

Now, making one nil, doesn't affect the other, which means that you can just pick up a rotating + translating object i.e. stop its translation, but still the ball will be rotating!

This clearly implies that even if you can't influence $E{rot}$ with $E{trans}$ or vice-versa.


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