Consider the system shown in the figure. A rope goes over a pulley. A mass, 𝒎, is hanging from the rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.

The pulley radius is 𝒓 and its mass moment of inertia is 𝑱. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural frequency of the system is

Correct Answer:

$${\omega }_n=\sqrt{\frac{k{r}^2}{J+m{r}^2}}$$

Step-by-Step Explanation:

1. Understanding the System Components and Variables:
Based on the provided figures, the physical system consists of:

• A hanging mass, m, which moves vertically.
• A spring with stiffness, k, anchored to a rigid wall on one end and attached to the rope on the other.
• A pulley with radius, r (labeled as R in the derivation diagram), and mass moment of inertia, J, pivoted at its center.
• An inextensible and massless rope that does not slip relative to the pulley.

2. Defining the Coordinates and Kinematic Relations:
Let us define the displacement of the system about its static equilibrium position:
Let $\theta$ be the angular rotation of the pulley from its static equilibrium position.
Since the rope does not slip on the pulley, the linear displacement $x$ of the mass m and the stretching/compression of the spring are directly related to the rotation of the pulley by the arc length formula:
$$x=r\theta$$
Differentiating this relation with respect to time gives the velocity:
$$\dot{x}=r\dot{\theta }$$

3. Formulation using the Energy Method:
For a conservative system vibrating about its static equilibrium position, the total mechanical energy (Kinetic Energy, K.E. + Potential Energy, P.E.) remains constant over time. Therefore, the time derivative of the total energy is zero:
$$\frac{d}{dt}(K.E.+P.E.)=0$$

The total Kinetic Energy (K.E.) of the system is the sum of the translational kinetic energy of the mass m and the rotational kinetic energy of the pulley:
$$K.E.=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}J{\dot{\theta }}^2$$
Substituting $\dot{x}=r\dot{\theta }$ into the kinetic energy equation:
$$K.E.=\frac{1}{2}m(r\dot{\theta }{)}^2+\frac{1}{2}J{\dot{\theta }}^2=\frac{1}{2}(J+m{r}^2){\dot{\theta }}^2$$

The Potential Energy (P.E.) stored in the spring when displaced by $x$ is:
$$P.E.=\frac{1}{2}k{x}^2=\frac{1}{2}k(r\theta {)}^2=\frac{1}{2}k{r}^2{\theta }^2$$

4. Deriving the Governing Equation of Motion:
Substituting the energy expressions into the conservation derivative:

Applying the chain rule for differentiation with respect to time:
$$\frac{1}{2}(J+m{r}^2)\cdot (2\dot{\theta }\breve{\theta })+\frac{1}{2}k{r}^2\cdot (2\theta \dot{\theta })=0$$
Simplifying this expression yields:

Since $\dot{\theta }$ is not identically zero for all time during vibration, the expression inside the brackets must be zero:
$$(J+m{r}^2)\breve{\theta }+k{r}^2\theta =0$$

5. Determining the Natural Frequency:
Rearranging the equation of motion into the standard harmonic oscillator form:
$$\breve{\theta }+\left(\frac{k{r}^2}{J+m{r}^2}\right)\theta =0$$
Comparing this with the general dynamic equation $\breve{\theta }+{\iota }_n^2\theta =0$, we obtain the natural frequency:
$${\omega }_n=\sqrt{\frac{k{r}^2}{J+m{r}^2}}$$