Why can we use this approximation in rotational work-kinetic energy theorem for small displacements?

[zenterix]

Homework Statement

In deriving the rotational work-kinetic energy theorem, the approximation is used that for small displacements

$$\frac{d\omega_z}{dt}d\theta=d\omega_z\frac{d\theta}{dt}=d\omega_z\omega_z\tag{6}$$

Why is this true?

Relevant Equations

Suppose we have a rigid body rotating about an axis and let $S$ be a point on this axis.

Define a coordinate system such that the $z$-axis coincides with the rotation axis.

If there is a net external torque on the system about $S$, then this net torque vector must have its direction on the $z$-axis (since the system is only rotating about this axis at all times).

This torque generates angular acceleration

$$\tau_{S,z}=I_S\alpha_z\tag{1}$$

where $I_S$ is moment of inertia about the axis of rotation and $\alpha_z$ is the z-component of angular acceleration, ie $\vec{\alpha}=\alpha_z\hat{k}$.

The force generating the torque does work as the rigid body rotates from an initial angle $\theta_i$ to $\theta_f$.

$$W_{rot}=\int_{\theta_i}^{\theta_f}\tau_{S,z}d\theta\tag{2}$$

$$=\int_{\theta_i}^{\theta_f} I_S\alpha_zd\theta\tag{3}$$

My question is about getting from this equation to a relationship with kinetic energy.

We have

$$dW_{rot}=I_S\alpha_zd\theta\tag{4}$$

$$=I_S\frac{d\omega_z}{dt}d\theta\tag{5}$$

The notes I am reading say that "in the limit of small displacements"

$$\frac{d\omega_z}{dt}d\theta=d\omega_z\frac{d\theta}{dt}=d\omega_z\omega_z\tag{6}$$

and so

$$dW_{rot}=I_Sd\omega_z\omega_z$$

and when we integrate this we easily reach the conclusion that $W_{rot}=\Delta K_{rot}$.

But why is (6) true?

 

[erobz]

I believe $\alpha$ can be reformulated via the chain rule in the integral as:

$$ \alpha = \frac{d \omega}{ d \theta} \frac{d \theta}{dt} $$

So if you plug that into the integral:

$$ W = I_s \int \frac{d \omega}{ d \theta} \frac{d \theta}{dt} d \theta = I_s \int \frac{d \omega}{ d \theta} \omega ~d \theta = I_s \int \omega ~d\omega $$

?

 

[zenterix]

I believe $\alpha$ can be reformulated via the chain rule in the integral as:

$$ \alpha = \frac{d \omega}{ d \theta} \frac{d \theta}{dt} $$

So if you plug that into the integral:

$$ W = I_s \int \frac{d \omega}{ d \theta} \frac{d \theta}{dt} d \theta = I_s \int \frac{d \omega}{ d \theta} \omega ~d \theta = I_s \int \omega ~d\omega $$

?

I get that the algebra works. I still wonder, as always with differentials, why it works.

$$\alpha_z=\frac{d\omega_z}{dt}=\theta''(t)$$

If we think of $\omega_z$ as a function of $\theta$ then sure, the chain rule says that

$$\alpha_z=\frac{d\omega_z}{d\theta}\frac{d\theta}{dt}$$

$\alpha_z$ is a rate of change relative to $t$.

What does it mean physically when we multiply it by $d\theta$ so that

$$\alpha_zd\theta=d\omega_z\frac{d\theta}{dt}$$

?

 

[renormalize]

What does it mean physically when we multiply it by $d\theta$ so that

$$\alpha_zd\theta=d\omega_z\frac{d\theta}{dt}$$

?

Multiply this equation by the moment-of-inertial $I$ to get:$$I\alpha\,d\theta=Id\omega\frac{d\theta}{dt}=Id\omega\,\omega=d\left(\frac{I\omega^{2}}{2}\right)$$ or using the definition of torque $\tau=I\alpha$:$$\tau d\theta=d\left(\frac{I\omega^{2}}{2}\right)$$This is just the differential form of the work-energy theorem for rotational motion: torque times infinitesimal angular displacement is equal to the differential change in the kinetic energy.

 

[erobz]

What does it mean physically when we multiply it by $d\theta$ so that

$$\alpha_zd\theta=d\omega_z\frac{d\theta}{dt}$$

?

Going to need a Mathematician and a Physicist for that. There is (I believe) debate around the Hyperreals, and what they mean formally in Mathematics vs. Physics. Way over my head.

 

[PeroK]

It's simply that for a small finite change:
$$\frac{\Delta \omega}{\Delta t}\Delta \theta = \Delta \omega \frac{\Delta \theta}{\Delta t}$$This then also applies to the differentials.

 

[kuruman]

Homework Statement: In deriving the rotational work-kinetic energy theorem, the approximation is used that for small displacements

$$\frac{d\omega_z}{dt}d\theta=d\omega_z\frac{d\theta}{dt}=d\omega_z\omega_z\tag{6}$$

Why is this true?

Here is another approach to answering your question that I hope will not make a mathematician cringe with horror. Start with finite displacements and assume constant angular acceleration. Using the SUVAT equations, we have $$\begin{align}LHS=\frac{\Delta \omega~\Delta \theta}{\Delta t}=\Delta \omega\left(\frac{\omega+\omega_0}{2}\Delta t\right)\times \frac{1}{\Delta t}=\frac{1}{2}\Delta \omega~(\omega+\omega_0). \end{align}$$Equation (1) is correct regardless of whether $\Delta t,$ and hence the displacement, are small or not.

If you consider infinitesimal changes $\Delta t \rightarrow dt\rightarrow 0,$ then $\Delta \omega\rightarrow d\omega$ and $(\omega+\omega_0)\rightarrow 2\omega.$ Thus, $$\lim_{\Delta t \rightarrow 0} LHS=\omega ~d\omega.$$ If the acceleration is not constant, Equation (1) is no longer appropriate for finite displacements. However, this does not invalidate the argument because we are considering infinitesimal differences. In other words, if you tell me "Yes, but the angular acceleration is not necessarily constant", I will tell you "OK, then I will consider an even smaller $\Delta t$ until the acceleration becomes constant over that interval." That's the beauty of differentials.

 

[PeroK]

Formally, the defining equation for differentials is:$$dA = \frac{dA}{dt}dt \ \text{and} \ dB = \frac{dB}{dt}dt$$In general, therefore:
$$dA\frac{dB}{dt}= \frac{dA}{dt}dt\frac{dB}{dt} = \frac{dA}{dt}dB $$This holds for any $A,B$ that are functions of $t$.