Two rotating doors, each 6.0 m long, are fixed to the same central axis of rotation, as shown (top view). When you push on one door with a constant 100 N, directed perpendicular from the face of the door and 50 cm from its outer edge, the rotating door system takes 8 s to complete a full revolution from rest. The doors can be modeled as thin rectangles (moments of inertia for thin rectangles, around two different axes, are shown for reference). Calculate the mass of the system.

• A. 21 kg
• B. 84 kg
• C. 104 kg
• D. 416 kg

Answer: (D)

Verified step by step guidance

1. Identify the torque applied to the door. Torque (τ) is calculated using the formula τ = F * r, where F is the force applied (100 N) and r is the distance from the axis of rotation (0.5 m).
2. Determine the angular acceleration (α) of the system. Use the kinematic equation for rotational motion: θ = 0.5 * α * t^2, where θ is the angular displacement (2π radians for one full revolution), and t is the time (8 s). Solve for α.
3. Use the relationship between torque and angular acceleration: τ = I * α, where I is the moment of inertia of the system. Rearrange to solve for I: I = τ / α.
4. Model the doors as thin rectangles. Use the moment of inertia formula for a rectangle rotating about an axis through its center: I = 1/12 * m * (A^2 + B^2), where A and B are the dimensions of the rectangle. Here, A = 6.0 m and B = 0.5 m.
5. Solve for the mass (m) of the system using the calculated moment of inertia (I) and the given dimensions of the doors. Rearrange the moment of inertia formula to find m: m = 12 * I / (A^2 + B^2).