Let's solve the problem step by step. --- ## **Given Data** - **Mass, \( m \):** \( 3\, \text{kg} \) - **Diameter, \( d \):** \( 400\, \text{mm} = .4\, \text{m} \) ⇒ **Radius, \( r \):** \( .2\, \text{m} \) - **Incline angle, \( \theta \):** \( 15^\circ \) - **Damping coefficient, \( c \):** \( 5\, \text{N}\cdot\text{s}/\text{m} \) - **Spring constant, \( k \):** \( 8\, \text{N}/\text{m} \) - **Initial displacement, \( x_ \):** \( .1\, \text{m} \) (above equilibrium) - **Initial velocity, \( v_ \):** \( -1.72\, \text{m/s} \) (downwards) - **No slipping:** Pure rolling condition --- ## **1. Model the System (Equation of Motion)** For a cylinder rolling **without slipping** on an incline with a spring and damper: ### **a. Free-Body Diagram** Forces along the incline (\( x \)-direction): - Spring force: \( -kx \) - Damping force: \( -c\dot{x} \) - Component of gravity: \( mg\sin\theta \) - Friction (ensures no slip, but does not do work since there is no slipping) ### **b. Rotational Inertia** Uniform cylinder: Moment of inertia about center, \( I_G = \frac{1}{2}mr^2 \) ### **c. Rolling Condition** No slip: \( a = r\alpha \) ⇒ \( \ddot{x} = r\ddot{\theta} \) ### **d. Equation of Motion** Apply Newton’s second law for translation and rotation: #### **Translation:** \[ m\ddot{x} = -kx - c\dot{x} + mg\sin\theta + F_f \] where \( F_f \) is the friction force **up the incline**. #### **Rotation:** \[ I_G\ddot{\theta} = F_f r \] But \( \ddot{x} = r\ddot{\theta} \Rightarrow \ddot{\theta} = \frac{\ddot{x}}{r} \): \[ I_G \frac{\ddot{x}}{r} = F_f r \implies F_f = \frac{I_G}{r^2} \ddot{x} \] #### **Combine:** Substitute \( F_f \): \[ m\ddot{x} = -kx - c\dot{x} + mg\sin\theta + \frac{I_G}{r^2} \ddot{x} \] \[ \left(m + \frac{I_G}{r^2}\right) \ddot{x} + c\dot{x} + kx = mg\sin\theta \] For a solid cylinder, \( I_G = \frac{1}{2}mr^2 \): \[ m + \frac{\frac{1}{2}mr^2}{r^2} = m + \frac{1}{2}m = \frac{3}{2}m \] So, \[ \frac{3}{2}m \ddot{x} + c\dot{x} + kx = mg\sin\theta \] Or, \[ \boxed{ \ddot{x} + \frac{2c}{3m}\dot{x} + \frac{2k}{3m}x = \frac{2g\sin\theta}{3} } \] --- ## **2. Substitute Numerical Values** Given: - \( m = 3\, \text{kg} \) - \( c = 5\, \text{N}\cdot\text{s}/\text{m} \) - \( k = 8\, \text{N}/\text{m} \) - \( g = 9.81\, \text{m}/\text{s}^2 \) - \( \theta = 15^\circ \) Plug into the equation: - \( \frac{2c}{3m} = \frac{2 \times 5}{3 \times 3} = \frac{10}{9} \approx 1.111\, \text{s}^{-1} \) - \( \frac{2k}{3m} = \frac{2 \times 8}{3 \times 3} = \frac{16}{9} \approx 1.778\, \text{s}^{-2} \) - \( \frac{2g\sin\theta}{3} = \frac{2 \times 9.81 \times \sin 15^\circ}{3} \) - \( \sin 15^\circ \approx .2588 \) - \( \frac{2 \times 9.81 \times .2588}{3} \approx \frac{5.075}{3} \approx 1.692\, \text{m/s}^2 \) So: \[ \boxed{ \ddot{x} + 1.111\,\dot{x} + 1.778\,x = 1.692 } \] --- ## **3. Solve the Differential Equation** ### **a. Homogeneous Solution** \[ \ddot{x} + 1.111\,\dot{x} + 1.778\,x = \] Characteristic equation: \[ r^2 + 1.111r + 1.778 = \] Discriminant: \[ \Delta = (1.111)^2 - 4(1.778) = 1.234 - 7.112 = -5.878 \] Negative, so **underdamped**. Roots: \[ r = \frac{-1.111 \pm \sqrt{-5.878}}{2} = -.5555 \pm j1.213 \] So, \[ x_h(t) = e^{-.5555 t} \left[ A \cos(1.213 t) + B \sin(1.213 t) \right] \] ### **b. Particular Solution** The particular solution for a constant right side is a constant: \[ x_p = \frac{1.692}{1.778} = .952 \] ### **c. General Solution** \[ \boxed{ x(t) = .952 + e^{-.5555 t} \left[ A \cos(1.213 t) + B \sin(1.213 t) \right] } \] --- ## **4. Initial Conditions** Given: - \( x() = x_ = .1 \) m (above equilibrium) - \( \dot{x}() = v_ = -1.72 \) m/s At \( t = \): \[ x() = .952 + A = .1 \implies A = .1 - .952 = -.852 \] Now, for \( \dot{x}() \): \[ \dot{x}(t) = e^{-.5555 t} [ -.5555 A \cos(1.213 t) - .5555 B \sin(1.213 t) - A \cdot 1.213 \sin(1.213 t) + B \cdot 1.213 \cos(1.213 t) ] \] At \( t = \): \[ \dot{x}() = [ -.5555 A + 1.213 B ] \] \[ -1.72 = -.5555(-.852) + 1.213 B \] \[ -1.72 = .473 + 1.213 B \] \[ 1.213 B = -1.72 - .473 = -2.193 \] \[ B = \frac{-2.193}{1.213} = -1.808 \] --- ## **5. **Final Solution** \[ \boxed{ x(t) = .952 + e^{-.5555 t} \Big[ -.852 \cos(1.213 t) - 1.808 \sin(1.213 t) \Big] } \] --- ### **Summary Table** | Parameter | Value | |-----------|-------| | \( m \) | \( 3\, \text{kg} \) | | \( r \) | \( .2\, \text{m} \) | | \( \theta \) | \( 15^\circ \) | | \( c \) | \( 5\, \text{N}\cdot\text{s}/\text{m} \) | | \( k \) | \( 8\, \text{N}/\text{m} \) | | \( x_ \) | \( .1\, \text{m} \) | | \( v_ \) | \( -1.72\, \text{m/s} \) | | \( x(t) \) | \( .952 + e^{-.5555 t} [ -.852 \cos(1.213 t) - 1.808 \sin(1.213 t) ] \) | --- ## **Interpretation** - The system is **underdamped** (oscillatory with decaying amplitude). - The cylinder's position as a function of time is given by the above formula. - The equilibrium position is at \( x = .952\, \text{m} \) **down the incline** from the spring's unstretched length. --- If you need a plot or further explanation, let me know!