## Step 1: Find the Moment of Inertia The moment of inertia \( I \) about the center involves: - disk: \(_{\text{disk}} = \frac{1}{2}MR^2 \) - Mass \( m \) at the rim: \( I_{m} = mR^2 \) - Mass \( 2m \) at the rim: \( I_{2m} = 2mR^2 \) **Total moment of inertia:** \[ I_{\text{total}} = \frac{1}{2}MR^2 + mR^2 + 2mR^2 = \frac{1}{2}MR^2 + 3mR^2 \] **Explanation:** We sum the moments of inertia for all masses about the disk’s center. --- ## Step 2: Write the Equation of Motion Using Torques - Let \(\theta\) be the angle from vertical for mass \(m\). - The torque due to gravity for each mass is: - For \(m\): \( -mgR \sin\theta \) - For \(2m\) (opposite side): \( +2mgR \sin\theta \) **Net torque:** \[ \tau = -mgR \sin\theta + 2mgR \sin(\theta + \pi) \] Since \(\sin(\theta + \pi) = -\sin\theta\), \[ \tau = -mgR \sin\theta - 2mgR \sin\theta = -3mgR \sin\theta \] From rotational dynamics: \[ I_{\text{total}} \ddot{\theta} = \tau \] \[ \left(\frac{1}{2}MR^2 + 3mR^2\right) \ddot{\theta} = -3mgR \sin\theta \] \[ \ddot{\theta} + \frac{3mg}{\left(\frac{1}{2}M + 3m\right)R} \sin\theta = \] **Explanation:** We apply Newton’s second law for rotation, sum the torques due to gravity, and write the equation for angular acceleration. --- ## **Final Answer** \[ \boxed{ \ddot{\theta} + \frac{3mg}{\left(\frac{1}{2}M + 3m\right)R} \sin\theta = } \] This is the required differential equation for \(\theta\).