Let's break down the problem step by step. --- ### Given Data - Initial angular velocity, \( \omega_ = 30. \) rad/s - Angular acceleration, \( \alpha = 40. \) rad/s\(^2\) (while the motor is connected) - Time interval, \( t_1 = 2.50 \) s (motor connected) - Additional angular displacement after disconnect, \( \theta_2 = 500 \) rad (while decelerating to stop) We will solve: #### a) **Angular displacement from \( t = \) to when motor was disconnected** **Step 1: Find final angular velocity at disconnect (\( \omega_1 \))** \[ \omega_1 = \omega_ + \alpha t_1 \] \[ \omega_1 = 30. + (40. \times 2.50) = 30. + 100. = 130. \ \text{rad/s} \] **Explanation:** This gives the angular velocity at the moment the motor is disconnected. --- **Step 2: Find angular displacement during this interval (\( \theta_1 \))** \[ \theta_1 = \omega_ t_1 + \frac{1}{2} \alpha t_1^2 \] \[ \theta_1 = (30. \times 2.50) + \frac{1}{2}(40.)(2.50^2) \] \[ \theta_1 = 75. + .5 \times 40. \times 6.25 \] \[ \theta_1 = 75. + 20. \times 6.25 \] \[ \theta_1 = 75. + 125. = 200. \ \text{rad} \] **Explanation:** This is the total angular displacement from \( t = \) to when the motor is disconnected. --- **Final Answer (a):** **Angular displacement from \( t = \) to disconnect:** \[ \boxed{200 \ \text{rad}} \] --- #### b) **Time required for the mass to come to a stop after disconnect** **Step 1: Use displacement and initial velocity after disconnect** We know: - Initial angular velocity after disconnect = \( \omega_1 = 130. \) rad/s - Angular displacement to stop = \( \theta_2 = 500 \) rad - Final angular velocity = (stopped) - Let angular acceleration after disconnect = \( \alpha_2 \) (unknown, negative) First, use kinematic equation: \[ \omega_f^2 = \omega_1^2 + 2\alpha_2 \theta_2 \] \[ = (130.)^2 + 2\alpha_2 \times 500 \] \[ = 16900 + 100\alpha_2 \] \[ 100\alpha_2 = -16900 \] \[ \alpha_2 = -16.9 \ \text{rad/s}^2 \] **Explanation:** This is the angular deceleration as the mass comes to a stop. --- **Step 2: Find time to stop (\( t_2 \)) after disconnect** \[ \omega_f = \omega_1 + \alpha_2 t_2 \] \[ = 130. + (-16.9)t_2 \] \[ 16.9t_2 = 130. \] \[ t_2 = \frac{130.}{16.9} \approx 7.69 \ \text{s} \] **Explanation:** This is the time needed for the mass to come to a stop after disconnect. --- **Final Answer (b):** **Time to stop after disconnect:** \[ \boxed{7.69 \ \text{s}} \] --- ### **Summary of Answers:** a) Angular displacement before disconnect: **200 rad** b) Time to stop after disconnect: **7.69 s**