Let's solve the problem step by step: ### **Given Data** - **Diameter of disk, \( d = 1.8 \) m** ⇒ **Radius, \( R = 0.9 \) m** - **Torque, \( \tau = 70.0 \) m·N** - **Total revolutions, \( N = 800 \)** - **Time, \( t = 25.0 \) s** - **Disk starts from rest** (\( \omega_0 = 0 \)) - **Moment of inertia for disk, \( I = \frac{1}{2}MR^2 \)** - **Find:** Mass \( M \) of the disk --- ### **Step 1: Find the Angular Displacement (\( \theta \))** 800 revolutions → radians: \[ \theta = 800 \times 2\pi = 1600\pi \ \text{radians} \] --- ### **Step 2: Find Angular Acceleration (\( \alpha \))** Use rotational kinematics: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \] Since disk starts from rest (\( \omega_0 = 0 \)): \[ \theta = \frac{1}{2} \alpha t^2 \implies \alpha = \frac{2\theta}{t^2} \] Plug in values: \[ \alpha = \frac{2 \times 1600\pi}{(25.0)^2} = \frac{3200\pi}{625} = \frac{3200 \times 3.1416}{625} = \frac{10053.1}{625} \approx 16.085 \ \text{rad/s}^2 \] --- ### **Step 3: Relate Torque, Moment of Inertia, and Angular Acceleration** \[ \tau = I \alpha \] \[ I = \frac{1}{2}MR^2 \] So, \[ \tau = \left(\frac{1}{2}MR^2\right)\alpha \] \[ M = \frac{2\tau}{R^2\alpha} \] --- ### **Step 4: Plug in All Values** - \( \tau = 70.0 \) N·m (mN = millinewton, but here it means meter-Newton) - \( R = 0.9 \) m - \( \alpha \approx 16.085 \) rad/s² \[ M = \frac{2 \times 70.0}{(0.9)^2 \times 16.085} = \frac{140.0}{0.81 \times 16.085} = \frac{140.0}{13.033} \approx 10.74 \ \text{kg} \] --- ### **Final Answer** \[ \boxed{M \approx 10.7\ \text{kg}} \] #### **Summary of Steps:** 1. Converted revolutions to radians. 2. Used kinematic equation to find angular acceleration. 3. Used torque and moment of inertia relation to solve for mass. 4. Plugged in values to calculate the mass.

10.7 kg

10.7 kg