11.4: Conservation of energy and potential energy

Like in Sec. 7.2, the analysis of rigid bodies using work and energy principles is facilitated when only conservative forces are acting on the system. For this situation, the work done can be described in terms of the change in potential energy since:

\[W_{\mathrm{tot}, \mathrm{cons}}=-\Delta V_{\mathrm{tot}} \tag{11.16} \label{11.16}\]

Combining this with Equation 11.15 we obtain, just like in Equation 7.13 that the total energy does not change:

Concept: Law of energy conservation for rigid bodies

\[\Delta V_{\mathrm{tot}}+\Delta T_{\mathrm{tot}}=0 \tag{11.17} \label{11.17}\]

And for a large number of objects, the conservation of energy equation between two times \(t_{1}\) and \(t_{2}\) is like Equation 7.34:

\[\sum_{i} T_{i}\left(t_{1}\right)+\sum_{i} V_{i}\left(t_{1}\right)=\sum_{i} T_{i}\left(t_{2}\right)+\sum_{i} V_{i}\left(t_{2}\right) \tag{11.18} \label{11.18}\]

We can still use the potential energy expression of forces derived in Sec. 7.3.

Example 11.1

Determine the speed of the yo-yo with radius \(R\) in Figure 11.1 after falling a distance \(\Delta h\) from rest.

Figure 11.1: A yo-yo \(A\) is dropped from rest. Determine its velocity \(\overrightarrow{\boldsymbol{v}}_{G}\) after falling a distance \(\Delta h\).

Solution

We first note that the kinetic \(T_{1}\) and potential energy \(V_{1}\) are zero at \(t=t_{1}\), since the yo-yo starts from rest. Then at \(t=t_{2}\) its gravitational potential energy has reduced to a value \(V_{2}=-m g \Delta h\). Since the force of the rope and the gravitational force are conservative forces, we can apply the law of energy conservation to determine the kinetic energy of the yo-yo at \(t_{2}\) :

\[\begin{align} T_{1}+V_{1} & =T_{2}+V_{2} \tag{11.19} \label{11.19}\\[4pt] 0+0 & =T_{2}-m g \Delta h \tag{11.20} \label{11.20}\\[4pt] T_{2} & =m g \Delta h \tag{11.21} \label{11.21}\end{align}\]

To solve the problem we need the constraint equation set by the rope. If the angle \(\phi\) increases, the rope releases from the yo-yo and the \(y\)-coordinate of the CoM of the yo-yo decreases:

\[\begin{align} -y_{G} & =\Delta h=\phi R+L_{0} \tag{11.22} \label{11.22}\\[4pt] v_{G, y} & =\dot{y}_{G}=-\omega R \tag{11.23} \label{11.23}\end{align}\]

where we took the time derivative of the constraint equation set by the rope of the yo-yo to relate it \(\mathrm{CoM}\) velocity to its angular velocity. Now we write the equation for the kinetic energy of the yo-yo \(T_{2}\), which is the sum of the translational and rotational kinetic energy:

\[\begin{align} T_{2} & =\frac{1}{2} m\left|\overrightarrow{\boldsymbol{v}}_{G}\right|^{2}+\frac{1}{2} I_{G} \omega^{2} \tag{11.24} \label{11.24}\\[4pt] & =\frac{1}{2} m v_{G, y}^{2}+\frac{1}{2}\left(\frac{1}{2} m R^{2}\right)\left(v_{G, y} / R\right)^{2} \tag{11.25} \label{11.25}\\[4pt] & =\frac{3}{4} m v_{G, y}^{2}=m g \Delta h \tag{11.26} \label{11.26}\\[4pt] \left|v_{G, y}\right| & =\sqrt{\frac{4}{3} g \Delta h} \tag{11.27} \label{11.27}\end{align}\]where we used that for a solid disk \(I_{G}=\frac{1}{2} m R^{2}\). Thus we obtain the velocity of the \(\mathrm{CoM}\) of the yo-yo. Note that the energy method does not give us the direction of the velocity. But in this case, no forces act in the \(x\) direction and it is straightforward to determine that it is downward \(\overrightarrow{\boldsymbol{v}}_{G}=-\sqrt{\frac{4}{3} g \Delta h} \hat{\boldsymbol{\jmath}}\). Note that this velocity is lower than that of a free falling disk because part of the gravitational work is converted to rotational kinetic energy.