Let's break down and answer each part of the problem based on the provided diagram and information: --- **Given:** - Pulley with radius \( r \), mass moment of inertia \( J \) - Spring with constant \( k \) - Hanging mass \( m \) - Displacement of mass: \( x(t) \) - Pulley rotation: \( \theta(t) \) - \( x(t) = r \theta(t) \) - External force: \( F(t) \) - Ignore gravity --- ### (1) **Kinetic Energy in Terms of \(\theta\):** - **Pulley rotational energy:** \( \frac{1}{2} J \dot{\theta}^2 \) - **Mass translational energy:** \( \frac{1}{2} m \dot{x}^2 \) - Since \( x = r\theta \), \( \dot{x} = r\dot{\theta} \): \[ KE = \frac{1}{2} J \dot{\theta}^2 + \frac{1}{2} m (r\dot{\theta})^2 = \frac{1}{2} J \dot{\theta}^2 + \frac{1}{2} m r^2 \dot{\theta}^2 = \frac{1}{2} (J + m r^2) \dot{\theta}^2 \] --- ### (2) **Potential Energy in Terms of \(\theta\):** - **Spring potential energy:** \( \frac{1}{2} k x^2 \) - Since \( x = r\theta \): \[ PE = \frac{1}{2} k x^2 = \frac{1}{2} k (r\theta)^2 = \frac{1}{2} k r^2 \theta^2 \] --- ### (3) **Equivalent Mass and Stiffness (W.r.t. \(\theta\)):** From the kinetic and potential energies: - **Equivalent mass:** \( J + m r^2 \) - **Equivalent stiffness:** \( k r^2 \) --- ### (4) **Equivalent Force in the Direction of \(\theta\):** The external force \( F(t) \) acts on the mass \( m \), causing a torque about the center of the pulley: - **Equivalent torque:** \( T_{eq} = r F(t) \) So, the equivalent force in the \(\theta\) direction (torque) is \( r F(t) \). --- ## **Summary of Answers** 1. **Kinetic energy:** \(\boxed{\frac{1}{2}(J + m r^2)\dot{\theta}^2}\) 2. **Potential energy:** \(\boxed{\frac{1}{2} k r^2 \theta^2}\) 3. **Equivalent mass:** \(\boxed{J + m r^2}\), **Equivalent stiffness:** \(\boxed{k r^2}\) 4. **Equivalent force (torque):** \(\boxed{r F(t)}\) Let me know if you need detailed derivations or further clarifications!