Mechanical systems and mechanisms: kinematics, dynamics, transmission and transformation
What the exam tests
Kinematics (motion without forces), dynamics (forces, torque, energy), and transmissions that change speed, torque or direction. Expect mobility calculations, velocity ratios, gear trains, belt/chain drives, cams and power screws.
Core rules
- Grübler/Kutzbach mobility: M = 3(n − 1) − 2j1 − j2, where n is the number of links including the frame, j1 lower pairs (pin, slider), j2 higher pairs (cam, gear contact).
- Four-bar: the Grashof condition s + l ≤ p + q gives at least one full rotation; if it is not satisfied, the linkage is a double-rocker.
- Slider-crank: x = r(1 − cosθ) + l − √(l² − r² sin²θ); quick-return occurs when the crank is offset.
- Gears: velocity ratio i = ω1/ω2 = z2/z1 = d2/d1; module m = d/z must match for meshing. Torque ratio is inverse to speed ratio if efficiency is neglected.
- Trains: i_total = product of driven teeth / product of driver teeth. Planetary: use the Willis equation (ω_s − ω_c)/(ω_r − ω_c) = −z_r/z_s.
- Belts: v = πdn/60; slip reduces ratio; chain drives keep an exact ratio.
- Dynamics: P = Tω, T = Iα, work–energy, efficiency η = P_out/P_in.
Worked example
A motor drives a 20-tooth gear at 1500 rpm. It meshes with a 60-tooth gear on the output shaft. Ratio i = 60/20 = 3, so output speed = 1500/3 = 500 rpm. If motor torque is 10 N·m and efficiency is 0.90, output torque = 10×3×0.90 = 27 N·m. Output power = Tω = 27×(500×2π/60) = 1414 W, close to input 10×157 = 1570 W times 0.90.
Traps
- Forgetting the frame as a link in mobility, or counting a gear mesh as a lower pair.
- Using pitch diameter and tooth number inconsistently.
- Applying the ideal torque ratio without efficiency.
- Sign errors in planetary gear trains.
- Confusing angular velocity (rad/s) with rpm.