Dynamic Analysis of Mechanisms

What dynamic analysis determines

Kinematics gives a mechanism's positions, velocities and accelerations. Dynamics relates that motion to applied loads, actuator torque and joint reactions. In inverse dynamics the motion is prescribed and the required forces are unknown. In forward dynamics the loads are prescribed and acceleration and subsequent motion must be solved with the constraints.

Planar rigid-link equations

For each moving link of mass m, mass centre G and mass moment of inertia I_G about the axis normal to its plane:

• Sum F_x = m a_Gx.
• Sum F_y = m a_Gy.
• Sum M_G = I_G alpha, with a consistent positive rotation direction.

Draw a separate free-body diagram for every link. Include weight, applied force, actuator couple and joint reactions. An ideal frictionless pin transmits two force components but no couple; the forces it applies to its two connected links are equal and opposite. A slider guide constrains transverse motion and supplies the corresponding reaction; include friction only when a friction model is specified.

Solve motion before forces

Determine the configuration and angular velocities from the linkage constraints. For two points A and B on a rigid link, use vector relations:

v_B = v_A + omega cross r_B/A

a_B = a_A + alpha cross r_B/A + omega cross (omega cross r_B/A).

The final term is centripetal and points towards A. Use the same relation to find a_G. A mechanism drawing alone is insufficient: lengths, configuration, input motion, masses, mass-centre locations, inertia values and external loads must be specified.

Worked example: a driven rod

A uniform slender rod has mass 4 kg and length 0.6 m. It rotates in a vertical plane about a fixed frictionless pin O at one end. At the instant considered the rod points horizontally right, its angular velocity is 3 rad/s counterclockwise and its angular acceleration is 2 rad/s² counterclockwise. Find the required applied couple T and the pin reaction. Take x right, y up and g = 9.81 m/s²; neglect other loads.

The centre is 0.3 m to the right of O. Its acceleration is

a_Gx = -omega²(0.3) = -2.7 m/s²,

a_Gy = alpha(0.3) = 0.6 m/s².

Hence O_x = 4(-2.7) = -10.8 N and O_y - 4(9.81) = 4(0.6), giving O_y = 41.64 N. The pin force points left and up.

Because O is fixed in an inertial frame, moments may also be taken directly about O using I_O = mL²/3 = 0.48 kg m²:

T - mg(L/2) = I_O alpha.

Therefore T = 11.772 + 0.96 = 12.732 N m counterclockwise.

Check about G: the pin contributes -0.3(41.64) = -12.492 N m and the weight has zero moment about G. Thus T - 12.492 = 0.240 N m, equal to I_G alpha = (mL²/12)(2). Both methods agree.

D'Alembert formulation

An equivalent bookkeeping method adds an inertia force -m a_G at G and an inertia couple -I_G alpha, then writes equilibrium equations. These are mathematical terms, not additional physical interaction forces. Do not include them and also put m a_G or I_G alpha on the right-hand side, which would count inertia twice.

Power check and limitations

For ideal constraints, actuator power plus power of external forces equals the rate of change of kinetic and potential energy. For the example, actuator power is T omega = 38.196 W. Gravitational potential energy rises at mg v_Gy = 35.316 W, and kinetic energy rises at I_O omega alpha = 2.88 W, confirming the result. Here v_Gy = omega L/2 = 0.9 m/s.

An energy equation is useful for input torque but generally cannot determine every joint reaction. Near a toggle position, force transmission and numerical conditioning require care. Real mechanisms may need friction, flexibility, impact or clearance models beyond the ideal rigid-link analysis.

Common errors

• Omitting centripetal acceleration even when angular acceleration is zero.
• Using I_G with moments about another point without the correct transport terms.
• Assuming constant input speed means every link has zero angular acceleration.
• Confusing mass moment of inertia (kg m²) with area moment of inertia (m⁴).

Quick check

If the rod's angular acceleration becomes zero at the same horizontal configuration, the required torque is still 11.772 N m to balance the gravitational moment. With omega still 3 rad/s, the horizontal pin force remains -10.8 N: zero angular acceleration does not eliminate centripetal acceleration.