1. **Problem Statement:** Determine the magnitude and direction of the normal force exerted on the disc by the bar (part a), and solve for the angular acceleration of the bar (part b). 2. **Given Data:** - Mass of disc, $m_d = 0.500$ kg - Mass of bar, $m_b = 1.500$ kg - Angular velocity of disc, $\omega = 9.00$ rad/s - Angular deceleration of disc, $\alpha_d = -100.0$ rad/s$^2$ - Radius of disc, $r = 0.065$ m (converted from 65 units assuming units are cm) - Length of bar, $L = 0.220$ m - Distance from A to B, $d_{AB} = 0.144$ m - Friction coefficients: $\mu_d = 0.400$, $\mu_b = 0.300$ - Moment of inertia of disc about centroid, $I_d = \frac{1}{2} m_d r^2$ - Moment of inertia of bar about centroid, $I_b = 6.08 \times 10^{-3}$ kgm$^2$ 3. **Part a: Normal Force on Disc by Bar** - The disc is decelerating with angular acceleration $\alpha_d$. - The friction force at point C opposes motion and affects the normal force. - The normal force $N$ acts perpendicular to the contact surface between disc and bar. **Step 1:** Calculate moment of inertia of disc: $$I_d = \frac{1}{2} m_d r^2 = \frac{1}{2} \times 0.500 \times (0.065)^2 = 0.00105625 \text{ kgm}^2$$ **Step 2:** Calculate torque due to friction and normal force at point C. **Step 3:** Use rotational dynamics equation for disc: $$\sum \tau = I_d \alpha_d$$ The torque due to normal force $N$ at radius $r$ is: $$\tau_N = N \times r$$ The friction force $f$ is related to $N$ by: $$f = \mu_d N$$ The friction force produces torque opposing rotation: $$\tau_f = f \times r = \mu_d N r$$ Total torque on disc: $$\tau = N r - \mu_d N r = N r (1 - \mu_d)$$ Set equal to rotational inertia times angular acceleration: $$N r (1 - \mu_d) = I_d \alpha_d$$ Solve for $N$: $$N = \frac{I_d \alpha_d}{r (1 - \mu_d)} = \frac{0.00105625 \times (-100)}{0.065 \times (1 - 0.400)} = \frac{-0.105625}{0.039} = -2.71 \text{ N}$$ Magnitude of normal force: $$|N| = 2.71 \text{ N}$$ Direction: Since $N$ is negative in calculation, it acts opposite to assumed positive direction, meaning it acts inward on the disc from the bar. 4. **Part b: Angular Acceleration of Bar $\alpha_b$** - The bar experiences torque due to friction and normal force at point C'. **Step 1:** Calculate torque on bar about pin B: Distance from B to C is $L = 0.220$ m. Torque due to normal force on bar: $$\tau_N = N \times L$$ Torque due to friction force on bar: $$\tau_f = \mu_b N \times L$$ Total torque on bar: $$\tau_b = N L - \mu_b N L = N L (1 - \mu_b)$$ **Step 2:** Use rotational dynamics for bar: $$\tau_b = I_b \alpha_b$$ Solve for $\alpha_b$: $$\alpha_b = \frac{\tau_b}{I_b} = \frac{N L (1 - \mu_b)}{I_b} = \frac{2.71 \times 0.220 \times (1 - 0.300)}{6.08 \times 10^{-3}} = \frac{2.71 \times 0.220 \times 0.7}{0.00608}$$ Calculate numerator: $$2.71 \times 0.220 \times 0.7 = 0.417$$ Calculate angular acceleration: $$\alpha_b = \frac{0.417}{0.00608} = 68.57 \text{ rad/s}^2$$ **Answer:** - a) Magnitude of normal force $N = 2.71$ N directed inward on the disc from the bar. - b) Angular acceleration of bar $\alpha_b = 68.57$ rad/s$^2$.