Question 5.14: A Pulley System Consider the pulley system shown in Figure 5...

The system has two mass elements, one translational block of mass and one pulley rotating about its fixed mass center. Their motions are related to each other by the geometric constraint, x=r \theta. The kinetic energy of the system is

T=T_{\text {block }}+T_{\text {pulley }}=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} I_{\mathrm{O}} \dot{\theta}^{2}.

The undeformed position, static equilibrium position, and dynamic position of the mass-spring system are shown in Figure 5.83. Choosing the static equilibrium position as the datum for the gravitational potential energy, we have

V_{\mathrm{g}}=-m g x.

The elastic potential energy is

V_{\mathrm{e}}=\frac{1}{2} k\left(x+\delta_{\mathrm{st}}\right)^{2}.

The total potential energy is

V=V_{\mathrm{g}}+V_{\mathrm{e}}=-m g x+\frac{1}{2} k x^{2}+\frac{1}{2} k \delta_{\mathrm{st}}^{2}+k x \delta_{\mathrm{st}}.

Because of the static equilibrium condition, m g=k \delta_{\text {st }} the expression of potential energy becomes

V=\frac{1}{2} k x^{2}+\frac{1}{2} k \delta_{\mathrm{st}}^{2}.

The total energy of the system is

T+V=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} I_{\mathrm{O}} \dot{\theta}^{2}+\frac{1}{2} k x^{2}+\frac{1}{2} k \delta_{\mathrm{st}}^{2},

for which the time derivative is

\frac{\mathrm{d}}{\mathrm{d} t}(T+V)=m \dot{x} \ddot{x}+I_{\mathrm{O}} \dot{\theta} \ddot{\theta}+k x \dot{x}.

Using the geometric constraint, x=r \theta, which implies that \dot{x}=r \dot{\theta} and \ddot{x}=r \ddot{\theta}, we can rewrite the above equation as

\frac{\mathrm{d}}{\mathrm{d} t}(T+V)=m r^{2} \dot{\theta} \ddot{\theta}+I_{\mathrm{O}} \dot{\theta} \ddot{\theta}+k r^{2} {\theta} \dot{\theta}.

Applying the principle of conservation of energy, we find

\left(I_{\mathrm{O}}+m r^{2}\right) \ddot{\theta}+k r^{2} \theta=0.