Pushing a door.

Top view of a door: drag where and in which direction you push; moment arm r⊥ and τ = rF sin θ update, the door swings by τ.

Open the door with τ = 6 N m using only 10 N: find the needed spot and angle; then show a push along the door gives zero torque.

Multiplying F by r instead of the perpendicular distance.

• Torque (moment of force) is the rotational analogue of force: τ = r × F, where r is the position of the point of application from the origin. A door turns most easily when pushed at right angles at its outer edge; a push on the hinge line does nothing.
• Magnitude τ = rF sin θ = r⊥F = rF⊥, where r⊥ is the perpendicular distance of the line of action from the origin (the moment arm) and F⊥ the component of F perpendicular to r.
• Torque is zero if F = 0, r = 0, or the line of action passes through the origin (θ = 0° or 180°). Its SI unit is N m and dimensions M L² T⁻², the same as work, but torque is a vector and work a scalar.
• Angular momentum of a particle about a point: l = r × p, magnitude l = rp sin θ = r⊥p.
• dl/dt = τ: the rate of change of angular momentum equals the torque, the rotational form of F = dp/dt.
• For a system, L = Σrᵢ × pᵢ and dL/dt = τ_ext. Internal torques cancel provided the forces between particles are equal, opposite and along the line joining them.
• If τ_ext = 0, L is constant: conservation of angular momentum, three scalar laws for Lₓ, L_y, L_z.
• A particle moving with constant velocity has constant angular momentum about any point: r sin θ is the fixed distance of its straight path from the point, and the direction of l does not change.
• A fast-spinning bicycle rim held by one string at one end of its axle does not fall; its angular momentum precesses about the string, turning about an axis perpendicular to both L and the torque.

Teaching this idea? Put the simulation on your own page or LMS with this code.