A Pulley System Consider the pulley system shown in Figure 5.82. A block of mass m is connected to a translational spring of stiffness k through a cable, which passes by a pulley with mass mp and radius r. The pulley rotates about the fixed mass center O. The moment of inertia of the pulley about

Question

A Pulley System

Consider the pulley system shown in Figure 5.82. A block of mass [latex]m[/latex] is connected to a translational spring of stiffness [latex]k[/latex] through a cable, which passes by a pulley with mass [latex]m_{\mathrm{p}}[/latex] and radius [latex]r[/latex]. The pulley rotates about the fixed mass center O. The moment of inertia of the pulley about its mass center is [latex]I_{\mathrm{O}}[/latex]. Use the energy method to derive the equation of motion.

Accepted Answer

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, [latex]x=r heta[/latex]. The kinetic energy of the system is

[latex]T=T_{ ext {block }}+T_{ ext {pulley }}=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} I_{\mathrm{O}} \dot{ heta}^{2}.[/latex]

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

[latex]V_{\mathrm{g}}=-m g x.[/latex]

The elastic potential energy is

[latex]V_{\mathrm{e}}=\frac{1}{2} k\left(x+\delta_{\mathrm{st}} ight)^{2}.[/latex]

The total potential energy is

[latex]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}}.[/latex]

Because of the static equilibrium condition, [latex]m g=k \delta_{ ext {st }}[/latex] the expression of potential energy becomes

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

The total energy of the system is

[latex]T+V=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} I_{\mathrm{O}} \dot{ heta}^{2}+\frac{1}{2} k x^{2}+\frac{1}{2} k \delta_{\mathrm{st}}^{2},[/latex]

for which the time derivative is

[latex]\frac{\mathrm{d}}{\mathrm{d} t}(T+V)=m \dot{x} \ddot{x}+I_{\mathrm{O}} \dot{ heta} \ddot{ heta}+k x \dot{x}.[/latex]

Using the geometric constraint, [latex]x=r heta[/latex], which implies that [latex]\dot{x}=r \dot{ heta}[/latex] and [latex]\ddot{x}=r \ddot{ heta}[/latex], we can rewrite the above equation as

[latex]\frac{\mathrm{d}}{\mathrm{d} t}(T+V)=m r^{2} \dot{ heta} \ddot{ heta}+I_{\mathrm{O}} \dot{ heta} \ddot{ heta}+k r^{2} { heta} \dot{ heta}.[/latex]

Applying the principle of conservation of energy, we find

[latex]\left(I_{\mathrm{O}}+m r^{2} ight) \ddot{ heta}+k r^{2} heta=0.[/latex]

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

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