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21.5: Work-Energy Theorem

Last updated

Jul 20, 2022
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- 21.4: Translation and Rotation of a Rigid Body Undergoing Fixed Axis Rotation
- 21.6: Worked Examples

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For a rigid body, we can also consider the work-energy theorem separately for the translational motion and the rotational motion. Once again treat the rigid body as a point-like particle moving with velocity $\overrightarrow{\mathbf{V}}_{c m}$ in reference frame O . We can use the same technique that we used when treating point particles to show that the work done by the external forces is equal to the change in kinetic energy

$\begin{array}{l} W_{\text {tans }}^{\text {ext }}=\int_{i}^{f} \overrightarrow{\mathbf{F}}^{\text {ext }} \cdot d \overrightarrow{\mathbf{r}}=\int_{i}^{f} \frac{d\left(m \overrightarrow{\mathbf{V}}_{c m}\right)}{d t} \cdot d \overrightarrow{\mathbf{R}}_{c m}=m \int_{i}^{f} \frac{d\left(\overrightarrow{\mathbf{V}}_{c m}\right)}{d t} \cdot \overrightarrow{\mathbf{V}}_{c m} d t \\ =\frac{1}{2} m \int_{i}^{f} d\left(\overrightarrow{\mathbf{V}}_{c m} \cdot \overrightarrow{\mathbf{V}}_{c m}\right)=\frac{1}{2} m V_{\mathrm{cm}, f}^{2}-\frac{1}{2} m V_{\mathrm{cm}, i}^{2}=\Delta K_{\text {trans }} \end{array} \nonumber$

For the rotational motion we go to the center of mass reference frame and we determine the rotational work done i.e. the integral of the z -component of the torque about the center of mass with respect to dθ as we did for fixed axis rotational work. Then

$\begin{array}{l} \int_{i}^{f}\left(\vec{\tau}_{\mathrm{cm}}^{\mathrm{ext}}\right)_{z} d \theta=\int_{i}^{f} I_{\mathrm{cm}} \frac{d \omega_{\mathrm{cm}, z}}{d t} d \theta=\int_{i}^{f} I_{\mathrm{cm}} d \omega_{\mathrm{cm}, z} \frac{d \theta}{d t}=\int_{i}^{f} I_{\mathrm{cm}} d \omega_{\mathrm{cm}, z} \omega_{\mathrm{cm}, z} \\ =\frac{1}{2} I_{\mathrm{cm}} \omega_{\mathrm{cm}, f}^{2}-\frac{1}{2} I_{\mathrm{cm}} \omega_{\mathrm{cm}, i}^{2}=\Delta K_{\mathrm{rot}} \end{array} \nonumber$

In Equation (21.5.2) we expressed our result in terms of the angular speed $\omega_{\mathrm{cm}}$ because it appears as a square. Therefore we can combine these two separate results, Equations (21.5.1) and (21.5.2), and determine the work-energy theorem for a rotating and translating rigid body that undergoes fixed axis rotation about the center of mass.

$\begin{array}{l} W=\left(\frac{1}{2} m V_{\mathrm{cm}, \mathrm{f}}^{2}+\frac{1}{2} I_{\mathrm{cm}} \omega_{\mathrm{cm}, f}^{2}\right)-\left(\frac{1}{2} m V_{\mathrm{cm}, \mathrm{f}}^{2}+\frac{1}{2} I_{\mathrm{cm}} \omega_{\mathrm{cm}, i}^{2}\right) \\ =\Delta K_{\text {tans }}+\Delta K_{\mathrm{rot}}=\Delta K \end{array} \nonumber$

Equations (21.4.1), (21.4.4), and (21.5.3) are principles that we shall employ to analyze the motion of a rigid bodies undergoing translation and fixed axis rotation about the center of mass.

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