Are there two work-energy theorems (rotational and translational) or just a single theorem for both?

Question

Suppose a body is able to rotate. If work is applied to it along a path $C$, the traditional work-kinetic energy theorem states that

$$W_{\mathrm{translational}} = \int_{C} \vec{F} \cdot d\vec{r} = \Delta \left(\frac{m v^2}{2}\right)$$

But there is also the equivalent principle relating work done by a torque and the resulting change in rotational kinetic energy,

$$W_\mathrm{rotational} = \int \vec{\tau} \cdot d\vec{\theta} = \Delta \left(\frac{I \omega^2}{2}\right)$$

My question is whether this two equations are valid separately for a body undergoing both translational and rotational motion, or is it only valid that the total work equals the total change in kinetic energy, i.e.,

$$W_{\mathrm{translational}} + W_\mathrm{rotational} = \Delta \left(\frac{m v^2}{2}\right) + \Delta \left(\frac{I \omega^2}{2}\right)$$

I hope the questio is clear.

Answer 1

Perhaps a derivation. For an extended body, let $\vec R$ describe the motion of the center of mass, and let $\vec r'$ be the position of a point on the rigid body with respect to the center of mass, such that the position of any point is given by $$\vec r = \vec R + \vec r'.$$ Next, we differentiate that expression to get, $$\vec v= \vec V + \vec v'$$ which gives the velocity of any point on the body. If calculate the torque $\vec \tau$ about the center of mass, we can express $\vec v'$ in terms of the angular velocity $\vec \omega$ as $\vec v' = \vec \omega \times \vec r'$ such that $$\vec v = \vec V + \vec \omega \times \vec r'.$$

Now, let us proceed to work. For a single particle, the net work is $\displaystyle \sum_{\rm time} \vec F \cdot \vec v \, \delta t$, so, for a collection of particles, the total work would be $\displaystyle \sum_{\rm body}\sum_{\rm time}\delta \vec F\cdot \vec v\, \delta t$ where $\delta \vec F$ is the net force on a particular element of the body.

We can translate this to an integral as follows, $$W = \int_{\rm time}\int_{\rm body} \mathrm d\vec F\cdot \vec v \, \mathrm dt$$

Note that to simplify the writing, we will first work on finding the $\mathrm dW$ over a small amount of time $\mathrm dt$, namely, $$\mathrm dW = \int_{\rm body} \mathrm d\vec F\cdot \vec v \, \mathrm dt$$ which can be rewritten as $$\mathrm dW = \int_{\rm body} \mathrm d\vec F\cdot \left( \vec V+\vec \omega \times \vec r' \right)\, \mathrm dt$$ using the velocity equation from above.

Note that $\mathrm d\vec F = \vec a \, \mathrm dm$ (Second Law) so we can write the above as, $$\mathrm dW = \int_{\rm body} \vec a \, \mathrm dm\cdot \vec V \, \mathrm dt + \int_{\rm body} \mathrm d \vec F \cdot\vec \omega \times \vec r'\, \mathrm dt$$ where the mixed product can be rewritten as $$\mathrm dW = \int_{\rm body} \vec a \, \mathrm dm\cdot \vec V \, \mathrm dt + \int_{\rm body} \vec \omega \cdot \vec r'\times \mathrm d\vec F\, \mathrm dt.$$ Realize that $\vec r' \times \mathrm d\vec F = \mathrm d\vec \tau$, the net torque acting on a a particle of the body.

Note that when integrating over the body, time is kept constant. Next, realize that $\vec a$ and $\vec V$ are the same over the whole body (since these describe the center of mass). Also, $\vec \omega$ does not change over the body. So, factor out these constants and you get,

$$\mathrm dW = \vec a\cdot \vec V \mathrm dt \int_{\rm body} \mathrm dm + \vec \omega \, \mathrm dt \cdot \int_{\rm body} \mathrm d\vec\tau .$$

Next, we see that $\displaystyle \int_{\rm body} \mathrm dm$ is simply the total mass $m$, whereas $\displaystyle \int_{\rm body} \mathrm d\vec \tau$ is the net torque $\vec \tau$, so, $$\mathrm dW = m\vec a\cdot \vec V\, \mathrm dt+\vec \tau \cdot \vec\omega \, \mathrm dt$$ which we can integrate over time to get the total work over the trajectory. Of course, the first term is just the change in kinetic energy of the center of mass, and the second term is the change in rotational kinetic energy, giving us our familiar and well-liked $$W=\Delta K_{\rm trans} + \Delta K_{\rm rot}.$$

From this, you can see that for a rigid body, the net work is indeed the sum of its translational and rotational changes in kinetic energies. It is one theorem, which in rotation only or translation only cases can be reduced to something more simple, but, fundamentally, is one big theorem that covers both simultaneously.

Answer 2

To make a clarification, I am unsure about relativistic effects, so I assume we are only considering non-relativistic speeds.

To answer your question, it is important to define our translational and rotational kinetic energy. For a body undergoing both translational and rotational kinetic energy, a suitable definition is as follows.

Translational Kinetic Energy: Refers to the kinetic energy expression obtained, when treating the object as a moving point mass, where the point is its center of mass.

Rotational Kinetic Energy: Refers to the kinetic energy expression obtained in an alternative inertial frame of reference, where the total momentum of the object is 0. (Or as we commonly say Centre of Mass frame).

Given these 2 explicit definitions, yes, both equations are separately valid for a body undergoing both motions.

Answer 3

What you call the "traditional work-energy theorem" is its application to a particle, or to an extended object where the direction of the net force acting on the object is through its center of mass (COM).

For extended objects where the net force does not act through the COM you must include the net torque and change in rotational kinetic energy (KE). Rotational and translational motion are independent motions, but, in the absence of dissipative forces, the total change in KE must be the sum of the rotational and translational KE.

Hope this helps.

Answer 4

The total kinetic energy of a system of particles is the sum of (a) the translational kinetic energy of the center of mass (CM) plus (b) the kinetic energy of motion about the CM.

Term (a) is ${1 \over 2} Mv^2$ where $M$ is the total mass and $v$ is the speed of the CM, as you state.

In general, term (b) is ${1 \over 2} \sum_{i}^{}m_iv_i^{'2}$ where $m_i$ is the mass of the $i_{th}$ particle and $v_i$ is the speed of the $i_{th}$ particle with respect to the CM.

A rigid body is a special case of a system of particles for which the distance between any two particles is fixed.

For a rigid body rotating about a fixed axis (more accurately rotating in a plane if the axis is moving), term (b) is ${1 \over 2}I \omega^2$ where $I$ is the moment if inertia about the axis and $\omega$ is the angular speed, as you state.

For motion of a rigid body about a point, term (b) is ${1 \over 2} \vec \omega \cdot \vec J$ where $\vec \omega$ is the angular velocity and $\vec J$ is the angular momentum, both taken about the point. Alternatively, term (b) is ${1 \over 2} \vec\omega \cdot \tilde I \cdot \vec \omega$ where $\tilde I$ is the inertia tensor.

The work done is the change in the total kinetic energy. For a rigid body, only the total external force does work. But, in general for a system of particles, internal forces also contribute to the work.

See a physics mechanics text for details, such as Classical Mechanics by Goldstein.