Let's solve each part step-by-step: --- ## **Given:** - Mass \(m\) slides on a vertical, circular wire hoop of radius \(R\). - Top half (\( \leq \theta \leq \pi\)) is frictionless. - Bottom half (\(\pi < \theta \leq 2\pi\)) has kinetic friction coefficient \(\mu_k\). - Mass is released from rest at the top (\(\theta = \)), reaches point \((R,)\) (\(\theta = \pi\)) with velocity \(v = \sqrt{\frac{gR}{2}}\). - Direction: Counterclockwise. --- ## **Part (a): Work Done by Friction** **Friction acts only for \(\pi < \theta \leq 2\pi\).** - **Friction Force Magnitude:** \(f = \mu_k N\), where \(N\) is the normal force. - **Path length for friction:** Half-circumference = \(\pi R\). ### **Find the Work Done by Friction:** \[ W_f = - f \cdot \text{distance} \] #### **Step 1: Find the Normal Force \(N\)** For a mass sliding on a vertical circle: - At the bottom half, the normal force \(N\) is not constant, but to **simplify** (as is standard for such problems), **use the average normal force** over the bottom half. - However, for a circular path, the friction force always acts tangentially, and its magnitude is always \(\mu_k mg\) (if the speed is not too high or too low). #### **Step 2: Work Done** The friction force always acts opposite to the direction of motion. Over the bottom half (\(\pi R\)): \[ W_f = -\mu_k mg \cdot (\pi R) \] --- ## **Part (b): Magnitude of the Average Force Due to Friction** \[ \text{Average force} = \mu_k mg \] This is because friction force is constant in magnitude throughout the bottom half. --- ## **Part (c): Angle \(\theta\) When Mass Stops** Let’s set up the **energy analysis**: ### **Step 1: Initial Energy at \(\theta = \pi\)** - At \(\theta = \pi\), position \((R,)\): - Kinetic energy: \(K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{gR}{2}\right) = \frac{mgR}{4}\) - Potential energy (taking the bottom as reference): \(U = mgR\) (since it's at height \(R\) above the bottom). ### **Step 2: As the mass slides further (from \(\theta = \pi\) to some \(\theta\)),** - Let it stop at point \(\theta_s\) (\(\pi < \theta_s < 2\pi\)). **Energy at stop:** - Kinetic energy = - Potential energy at \(\theta_s\): \(U = mgR(1 + \cos\theta_s)\) - This comes from the vertical height above the bottom: \(y = R\cos\theta\), so height above the bottom is \(R(1 + \cos\theta)\). **Work by friction:** Over arc length \(s = R(\theta_s - \pi)\): \[ W_f = -\mu_k mg \cdot R (\theta_s - \pi) \] ### **Step 3: Apply Conservation of Energy** \[ \text{Initial Energy} = \text{Final Energy} + \text{Work by friction} \] \[ \left[\frac{mgR}{4} + mgR\right] = [mgR(1 + \cos\theta_s)] + [-\mu_k mg R (\theta_s - \pi)] \] \[ \frac{5mgR}{4} = mgR(1 + \cos\theta_s) - \mu_k mg R (\theta_s - \pi) \] Divide both sides by \(mgR\): \[ \frac{5}{4} = 1 + \cos\theta_s - \mu_k(\theta_s - \pi) \] \[ \frac{5}{4} - 1 = \cos\theta_s - \mu_k(\theta_s - \pi) \] \[ \frac{1}{4} = \cos\theta_s - \mu_k(\theta_s - \pi) \] \[ \cos\theta_s = \frac{1}{4} + \mu_k(\theta_s - \pi) \] **This is the equation for \(\theta_s\).** --- ## **Final Answers** ### **(a) Work done by friction** \[ \boxed{W_f = -\mu_k mg \pi R} \] --- ### **(b) Magnitude of the average force due to friction** \[ \boxed{F_{\text{friction}} = \mu_k mg} \] --- ### **(c) Angle \(\theta\) when mass stops** \[ \boxed{ \cos\theta_s = \frac{1}{4} + \mu_k(\theta_s - \pi) } \] You may solve numerically for \(\theta_s\) for a given \(\mu_k\). --- **Let me know if you'd like a specific value for \(\theta_s\) with a given \(\mu_k\)!**