Question 6 (paper Question VI) — Natural Frequency of a Disk-Spring-Block System (Part B · Dynamics, 20 marks): 04-BS-3, May 2015, Q6

Question 6 (paper Question VI) — Natural Frequency of a Disk-Spring-Block System (Part B · Dynamics, 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Uniform disk pinned (frictionless) at its centre O: radius r = 0.75 ft, weight 15 lb_f. A cord wraps the disk at radius r without slipping, connecting on one side to a spring (k = 80 lb_f/ft) and on the other side to a hanging block A (weight 3 lb_f).

Given data
Quantity	Value
Disk radius r	0.75 ft
Disk weight	15 lb_f
Block A weight	3 lb_f
Spring constant k	80 lb_f/ft
g	32.2 ft/s²

Find. The natural frequency and period of vibration of the system.

Figure 6 — disk pinned at O, cord (radius r, no slip) to a spring on one side and to block A on the other.

Approach. Use the energy method: write the kinetic energy of the disk (rotating about the fixed pin O) plus block A (moving with the cord), and the elastic potential energy of the spring, both in terms of a single coordinate θ (disk rotation, with $x=r\theta$ common to both cord segments); apply Lagrange's equation to get the equation of motion and read off ω_n.

Mass moment of inertia of the disk about the pin O. $$I_O = \tfrac12 m_{disk}r^2, \qquad m_{disk}=\dfrac{15}{32.2}=0.4658\ \text{slug} \;\Rightarrow\; I_O = 0.1310\ \text{slug}\cdot\text{ft}^2$$
Kinetic energy in terms of θ. Both A and the spring's cord end move at $r\dot\theta$ (no slip, common radius r on opposite sides): $$T = \tfrac12 I_O\dot\theta^2 + \tfrac12 m_A(r\dot\theta)^2 = \tfrac12\left(I_O+m_Ar^2\right)\dot\theta^2, \qquad m_A=\dfrac{3}{32.2}=0.09317\ \text{slug}$$
Elastic potential energy and equation of motion. Measuring θ from static equilibrium (gravity's constant moment cancels the static spring pre-stretch, standard for a vertical spring-mass system), $V=\tfrac12 k(r\theta)^2$. Lagrange's equation gives: $$\left(I_O+m_Ar^2\right)\ddot\theta + kr^2\theta = 0 \;\Rightarrow\; \omega_n^2 = \dfrac{kr^2}{I_O+m_Ar^2}$$
Evaluate ω_n. Substituting $I_O=\tfrac12 m_{disk}r^2$, the $r^2$ cancels — a useful check: $$\omega_n = \sqrt{\dfrac{k}{\tfrac12 m_{disk}+m_A}} = \sqrt{\dfrac{80}{0.2329+0.09317}} = \sqrt{245.3}=\boxed{15.66\ \text{rad/s}}$$
Frequency and period. $$f_n = \dfrac{\omega_n}{2\pi} = \boxed{2.49\ \text{Hz}}, \qquad T_n=\dfrac1{f_n}=\boxed{0.401\ \text{s}}$$

Results
Quantity	Value
Natural circular frequency ω_n	15.66 rad/s
Natural frequency f_n	2.49 Hz
Period of vibration T_n	0.401 s

Topic: Free vibration of a rigid body via the energy (Lagrange) method.

Key relations: $I_O=\tfrac12 mr^2$ for a uniform disk about its centre; energy method $\dfrac{d}{dt}\left(\dfrac{\partial T}{\partial\dot\theta}\right)+\dfrac{\partial V}{\partial\theta}=0$ gives $\omega_n^2=k_{eff}/m_{eff}$.

Why this works: because the disk is pinned at a fixed point, its whole kinetic energy is rotational; converting the hanging block's translational motion to the same angular coordinate θ (via the no-slip cord) lets the whole system be reduced to one effective mass-spring oscillator.

Common pitfall: forgetting to add the block's own inertia term $m_Ar^2$ to $I_O$ (using the disk's inertia alone), or mistakenly leaving r in the final answer for ω_n instead of recognizing that it cancels for this configuration.

Source: Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Ch. 22 (Vibrations).

Question 6 of 6: Question 6 (paper Question VI) — Natural Frequency of a Disk-Spring-Block System (Part B · Dynamics, 20 marks)

Question 6 (paper Question VI) — Natural Frequency of a Disk-Spring-Block System (Part B · Dynamics, 20 marks)