Kinematics and Dynamics I: Plane Mechanisms, Velocity and Acceleration Analysis, Dynamic Analysis of Linkages, Gears and Gear Trains, and Balancing
1. Plane mechanisms: degrees of freedom, inversions and instantaneous centres
A plane mechanism of n links (the fixed link included), j₁ lower pairs (one degree of freedom each: revolute or prismatic) and j₂ higher pairs has, by Grübler’s criterion, M = 3(n − 1) − 2j₁ − j₂ degrees of freedom. A four-bar chain has 3 × 3 − 2 × 4 = 1; a Watt six-bar chain of six links and seven revolute pairs has 3 × 5 − 2 × 7 = 1; a five-bar with five revolute pairs has 2. A mechanism needs M = 1 to be driven by one input; M = 0 is a structure and M < 0 a statically indeterminate one.
Grashof’s law: in a four-bar chain, if the shortest plus the longest link is no more than the sum of the other two, s + l ≤ p + q, some link can rotate fully relative to the others. Then fixing the link adjacent to the shortest gives a crank–rocker, fixing the shortest gives a double crank (drag link), and fixing the link opposite to it gives a double rocker; if s + l > p + q, every inversion is a double rocker. With links of 20, 45, 50 and 60 mm, s + l = 80 ≤ 95, so the chain is Grashof, and with the 20 mm link fixed it is a double crank. The slider-crank chain has four inversions: the reciprocating engine, the Whitworth quick-return and crank-and-slotted-lever mechanisms, the oscillating cylinder, and the hand pump.
The instantaneous centre of a body in plane motion is the point of zero velocity. A mechanism of N links has N(N − 1)/2 of them, one for each pair of links, so six links have 15 and four have 6; Kennedy’s theorem says the three instantaneous centres of any three bodies lie on one straight line. Any point’s velocity is then ω times its distance from the centre, at right angles to the line joining them.
2. Velocity and acceleration analysis
The relative-velocity method writes v_B = v_A + vB/A, with vB/A = ω × rB/A perpendicular to the link AB and of magnitude ω_AB·AB, and closes a velocity polygon. For the slider-crank of crank radius r and connecting-rod length l = nr, the slider’s velocity is v = ωr(sin θ + sin 2θ/2n); at θ = 90° it is exactly ωr, so 10 m/s for r = 0.1 m and ω = 100 rad/s, whatever n is. The slider’s acceleration is a = ω²r(cos θ + cos 2θ/n), greatest at the dead centre θ = 0, ω²r(1 + 1/n), which is why the inertia force of a piston is greatest there.
In the acceleration analysis every point on a rotating link has a normal (centripetal) component ω²r towards the centre and a tangential component αr. When a point also slides along a rotating link, a third term appears, the Coriolis acceleration a_c = 2ω v_rel, perpendicular to the link, in the direction of v_rel rotated by 90° in the sense of ω. A block sliding outward along a link at 1.5 m/s relative to it, while the link turns at 8 rad/s, has a Coriolis component 2 × 8 × 1.5 = 24 m/s². It is absent in a mechanism with no sliding on a moving link, and it is why the acceleration polygon of a quick-return mechanism differs from that of a slider-crank.
3. Dynamic analysis of linkages and the flywheel
D’Alembert’s principle turns a dynamics problem into a statics one: add an inertia force F_i = −m a_G, acting through the centre of mass, and an inertia torque T_i = −I_G α, to every moving link, and the link is in equilibrium under the applied, constraint and inertia loads. The two can be combined into a single force displaced by h = I_G α/(m a_G) from the centre of mass. Working from the output link back towards the input gives the torque the driver must supply and the forces on every pin, which is what sizes the pins and the frame.
A connecting rod is replaced for the analysis by a dynamically equivalent system of two point masses, which requires the same total mass, the same centre of mass and the same moment of inertia. Because the torque of a reciprocating engine fluctuates, a flywheel stores the excess energy and returns it. Its size follows from the coefficient of fluctuation of speed C_s = (ω_max − ω_min)/ω_mean and the maximum fluctuation of energy ΔE read from the turning-moment diagram: ΔE = I ω_mean² C_s. For ΔE = 400 J, N = 300 rpm (ω = 31.42 rad/s) and C_s = 0.02, I = 400/(986.96 × 0.02) = 20.3 kg·m².
4. Gears and gear trains
The law of gearing: for a constant velocity ratio the common normal at the point of contact must always pass through the pitch point on the line of centres, dividing it inversely as the angular velocities. The involute profile satisfies it, and keeps the ratio constant even if the centre distance varies slightly. Gears that mesh share the same module m = d/z (pitch diameter over teeth) and pressure angle, usually 20°, so the centre distance is a = m(z₁ + z₂)/2 and the velocity ratio is ω₁/ω₂ = z₂/z₁. Modules of 4 mm on 20 and 60 teeth give a = 4 × 80/2 = 160 mm. The contact ratio, the average number of tooth pairs in mesh, must exceed 1, and preferably 1.2. Interference, the tip of one tooth cutting into the root of the other, is avoided by at least about 17–18 teeth on the pinion of a 20° full-depth system (z_min = 2/sin²φ = 17.1).
In a simple train each shaft carries one gear, the intermediate gears (idlers) change the direction of rotation and not the ratio, and the overall ratio is the ratio of the last to the first. In a compound train two gears sit on one shaft and the ratio is the product of the tooth ratios: 20 → 40 on the first pair and 15 → 60 on the second gives (40/20) × (60/15) = 8, so 1200 rpm becomes 150 rpm. In an epicyclic train an arm carries planets that mesh with a sun and a ring; the speeds follow from the relative-speed table, (N_ring − N_arm)/(N_sun − N_arm) = −Z_sun/Z_ring for a sun–planet–ring train. With the ring fixed, the sun turns (1 + Z_ring/Z_sun) times the arm: for 20 and 80 teeth, 5.
5. Balancing of rotating and reciprocating masses
A rotating mass m at radius r produces a centrifugal force mω²r that turns with the shaft. It is balanced when the vector sum of all the mr products is zero (static balance), and, for masses in different planes, when the moments of those products about a reference plane also sum to zero (dynamic balance, the couple condition). A single 5 kg mass at 0.2 m is balanced by a mass at the diametrically opposite side: m_b r_b = 5 × 0.2, so 10 kg at 0.1 m. For several masses, close the polygon of mr vectors; for several planes, close the couple polygon too, with the balancing masses placed in two chosen planes.
In a reciprocating engine the inertia force of the piston mass m along the line of stroke is, with n = l/r, primary F_p = mω²r cos θ (once per revolution) and secondary F_s = mω²r cos 2θ/n (twice per revolution, smaller by 1/n). A rotating balance mass can cancel the primary force along the stroke only by producing an equal force at right angles, so a single cylinder is at best partially balanced: balancing a fraction c of the reciprocating mass leaves (1 − c) mω²r cos θ along the stroke and c mω²r sin θ across it. Multi-cylinder engines balance one another: an in-line four-cylinder engine with cranks 0°–180°–180°–0° has its primary forces and couples balanced, and unbalanced secondary forces. In a locomotive the unbalanced part of the balancing mass gives hammer blow, a variation in wheel load, and the along-track unbalance a swaying couple and variation of tractive effort.
Key takeaways
- Grübler: M = 3(n − 1) − 2j₁ − j₂; Grashof: s + l ≤ p + q; N(N − 1)/2 instantaneous centres; Kennedy: three centres of three bodies are collinear.
- Slider-crank velocity is ωr at θ = 90°; Coriolis acceleration 2ω v_rel needs both rotation and sliding.
- D’Alembert: F_i = −m a_G, T_i = −I α; flywheel I = ΔE/(ω² C_s).
- Module m = d/z; a = m(z₁ + z₂)/2; compound train ratio is the product of tooth ratios; idlers reverse direction only; ring fixed: sun/arm = 1 + Z_r/Z_s.
- Rotating: Σmr = 0 and Σmrx = 0; reciprocating: primary mω²r cos θ, secondary mω²r cos 2θ/n, only partly balanced in one cylinder.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A plane mechanism of six links, including the fixed link, is connected by seven revolute pairs (Watt’s six-bar chain). Its degree of freedom by Grübler’s criterion is ____.
Numerical answer — type the value.
Show answer
Answer: 1
M = 3(n − 1) − 2j₁ − j₂ = 3 × 5 − 2 × 7 − 0 = 1. Counting all six links, 3 × 6 − 14 = 4, forgets to subtract the fixed link’s three degrees of freedom.A four-bar chain has links of 20 mm, 45 mm, 50 mm and 60 mm. With the 20 mm link fixed, the mechanism is a:
Show answer
Answer: A — double-crank (drag-link) mechanism
s + l = 20 + 60 = 80 ≤ p + q = 95, so Grashof’s condition holds, and fixing the shortest link makes both adjacent links full cranks. Fixing a link next to the shortest gives a crank–rocker, and the opposite link a double rocker.The number of instantaneous centres of velocity in a plane mechanism of six links is ____.
Numerical answer — type the value.
Show answer
Answer: 15
N(N − 1)/2 = 6 × 5/2 = 15, one for every pair of links. Six is the count for a four-link mechanism, 4 × 3/2.In a slider-crank mechanism the crank has radius 0.1 m and turns at 100 rad/s. When the crank is at 90° to the line of stroke, the velocity of the slider is ____ m/s.
Numerical answer — type the value.
Show answer
Answer: 10
v = ωr(sin θ + sin 2θ/2n); at θ = 90°, sin θ = 1 and sin 2θ = 0, so v = ωr = 100 × 0.1 = 10 m/s, independent of the rod length. At the dead centres the velocity is zero.A block slides outward along a link at a velocity of 1.5 m/s relative to the link, while the link rotates at a constant 8 rad/s. The Coriolis component of the block’s acceleration is ____ m/s².
Numerical answer — type the value.
Show answer
Answer: 24
a_c = 2ω v_rel = 2 × 8 × 1.5 = 24 m/s², perpendicular to the link. Omitting the factor 2 gives 12, and ω²r would be the (separate) centripetal component.According to Kennedy’s theorem, the three instantaneous centres of three bodies in relative plane motion:
Show answer
Answer: A — lie on one straight line
The theorem states that they are collinear, which is what lets an unknown centre be found as the intersection of two lines through known ones. They coincide only in special cases such as pure rolling contact.A flywheel must limit the fluctuation of speed to a coefficient of 0.02 about a mean speed of 300 rpm. The maximum fluctuation of energy per cycle is 400 J. The moment of inertia required, correct to one decimal place, is ____ kg·m².
Numerical answer — type the value.
Show answer
Answer: 20.3
ω = 2π × 300/60 = 31.416 rad/s, ω² = 986.96, and I = ΔE/(ω² C_s) = 400/(986.96 × 0.02) = 20.26 ≈ 20.3 kg·m². Using N = 300 directly as ω would give a value 100 times too small.The law of gearing states that, for a constant velocity ratio, the common normal at the point of contact of the teeth must:
Show answer
Answer: A — always pass through the pitch point on the line of centres
The pitch point divides the line of centres inversely as the angular velocities, so a fixed pitch point means a fixed ratio. The involute profile produces a common normal of fixed direction, the line of action, that cuts the line of centres at one point.Two spur gears of module 4 mm, with 20 and 60 teeth, mesh with standard centre distance. The centre distance is ____ mm.
Numerical answer — type the value.
Show answer
Answer: 160
a = m(z₁ + z₂)/2 = 4 × (20 + 60)/2 = 160 mm; the pitch diameters are 80 and 240 mm, whose radii sum to 40 + 120. Using m(z₁ + z₂) = 320 confuses the sum of diameters with the sum of radii.In a compound gear train, a 20-tooth gear drives a 40-tooth gear; on the same shaft as the 40-tooth gear a 15-tooth gear drives a 60-tooth gear. The input speed is 1200 rpm. The output speed is ____ rpm.
Numerical answer — type the value.
Show answer
Answer: 150
The overall ratio is (40/20) × (60/15) = 2 × 4 = 8, so the output speed is 1200/8 = 150 rpm, in the same direction after two meshes. Dividing by 40/20 only gives 600.In an epicyclic gear train the sun has 20 teeth and the ring has 80 teeth. With the ring held fixed, the sun is the input and the arm is the output. The speed ratio (sun speed divided by arm speed) is ____.
Numerical answer — type the value.
Show answer
Answer: 5
(N_ring − N_arm)/(N_sun − N_arm) = −Z_sun/Z_ring. With N_ring = 0: −N_arm/(N_sun − N_arm) = −20/80, so N_sun − N_arm = 4N_arm and N_sun/N_arm = 5 = 1 + Z_ring/Z_sun. The value 4 is the difference, not the ratio.In a simple gear train, an intermediate (idler) gear is inserted between the driver and the driven gear. The effect is that:
Show answer
Answer: A — the speed ratio is unchanged and the direction of rotation of the driven gear is reversed
The idler’s tooth ratio appears once as z_idler/z_driver and once, inverted, as z_driven/z_idler, so it cancels: the ratio is z_driven/z_driver. The idler adds one more external mesh, reversing the sense; it also bridges a large centre distance.A rotating mass of 5 kg is fixed at a radius of 0.2 m on a shaft. It is balanced by a second mass placed diametrically opposite at a radius of 0.1 m in the same plane. The balancing mass is ____ kg.
Numerical answer — type the value.
Show answer
Answer: 10
Balance requires m_b r_b = m r = 5 × 0.2 = 1 kg·m, so m_b = 1/0.1 = 10 kg. The same rotating speed cancels from both sides. A mass of 5 kg would balance only at the equal radius of 0.2 m.In a quick-return mechanism the cutting stroke takes 220° of crank rotation and the return stroke takes 140°. The ratio of the time of the cutting stroke to that of the return stroke, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.57
The crank turns at constant speed, so times are proportional to angles: 220/140 = 1.571 ≈ 1.57. A ratio above 1 is what makes the return quick. The angles add to 360°.Which of the following statements about the balancing of a reciprocating engine are true?
Show answer
Answer: A — The primary inertia force has amplitude mω²r and acts along the line of stroke; B — The secondary force has twice the frequency of rotation and amplitude mω²r/n, with n = l/r; D — Partial balancing reduces the unbalanced force along the stroke but adds an unbalanced force at right angles to it
(A) F_p = mω²r cos θ. (B) F_s = mω²r cos 2θ/n at twice the shaft frequency. (C) False: a rotating mass produces a force of constant magnitude rotating with the shaft, which cancels the varying along-stroke force only by adding an equal one across it. (D) That is the trade-off of partial balancing.Which of the following statements about gears are true?
Show answer
Answer: A — An involute profile keeps the velocity ratio constant even if the centre distance changes slightly; B — Meshing gears must have the same module and the same pressure angle; C — The module of a gear is the pitch diameter divided by the number of teeth
(A) The base circles fix the ratio, so a small change of centre distance only changes the pressure angle. (B) Otherwise the teeth cannot mesh. (C) m = d/z. (D) False: an idler cancels out of the ratio and only reverses the direction.