Planar kinetics of rigid bodies: force and acceleration

With rigid-body kinematics in hand, we can now ask the kinetic questions that matter most in machine design. How much torque does a motor need to bring a conveyor drum up to speed in three seconds? Will a tall load tip over when a shuttle brakes, or slide first? What force does a bearing feel when a crank spins? Will a roll of film roll down a ramp or skid? Each needs two equations working together: the second law for the motion of the centre of mass, and a moment equation for the rotation. This lesson develops both, introduces the mass moment of inertia that governs rotation, and applies them to the three kinds of planar motion.

Mass moment of inertia

A body's mass measures how hard it is to accelerate in a straight line. Its mass moment of inertia measures how hard it is to spin up about an axis. For a body made of mass elements \(dm\) at distances \(r\) from the axis,

Mass far from the axis counts much more than mass near it, because of the square. A flywheel is built with a heavy rim and a light hub for exactly this reason: it stores as much rotational inertia as possible per kilogram. The units are kilogram-square metres in SI and slug-square feet in US customary units.

The radius of gyration \(k\) is defined by \(I = m k^2\): it is the distance from the axis at which all the mass could be concentrated to give the same moment of inertia. Manufacturers of motors, gearboxes and pulleys often list inertia directly, but older catalogues and some industries quote the related quantity \(W k^2\), or \(GD^2\), so check the definition before using a number.

The parallel-axis theorem gives the moment of inertia about any axis parallel to one through the centre of mass \(G\), a distance \(d\) away:

The rod results in the table illustrate it: \(mL^2/12 + m(L/2)^2 = mL^2/3\). For composite bodies, such as a drum made of a shell, two end discs and a shaft, add the moments of inertia of the parts about the same axis, using the parallel-axis theorem for any part whose centre of mass is not on the axis, and subtracting holes.

Equations of motion for a rigid body in plane motion

A rigid body is a system of particles, so Lesson 7 already gives the first equation: the resultant external force equals the total mass times the acceleration of the centre of mass. Summing moments of the particle equations about \(G\), and using the fact that the particles keep fixed distances, gives the second:

Three scalar equations for a body in plane motion. The moment equation must be written about the centre of mass \(G\), or about a fixed axis of rotation, unless the kinetic moment of \(m\mathbf{a}_G\) is included. In general, for moments about any point \(P\),

This form is often the most efficient, because choosing \(P\) at the intersection of two unknown forces removes them from the equation. It is easiest to apply by drawing the kinetic diagram: the body with a vector \(m\mathbf{a}_G\) at \(G\) and a couple \(I_G\alpha\). The moments of the external forces on the free-body diagram about any point equal the moments of \(m\mathbf{a}_G\) and \(I_G\alpha\) about the same point on the kinetic diagram.

Translation

When a body translates, \(\alpha = 0\), so \(\sum M_G = 0\): the forces have no net moment about the centre of mass, even though the body accelerates. This is enough to decide when a load on an accelerating vehicle tips over. The friction force at the base acts at floor level, below the centre of mass, and its moment tends to tip the load forward when the vehicle brakes. The load stays upright only if the normal force can shift far enough forward to balance that moment. When the normal force reaches the front edge of the base, the load is on the point of tipping.

The comparison turns into a simple rule: tipping governs whenever \(h > b/\mu_s\), that is, when the load is tall compared with its base and the friction is good. Good friction is not always good news: a grippy deck mat that stops totes sliding can make tall ones tip instead. The program below tabulates both limits for a range of load heights and works out an allowable braking rate for a shuttle that must handle every load up to 0.45 m tall, with a safety factor of 1.5.

# A load stands on a shuttle deck that brakes at deceleration a.
# It slides if a > mu_s g, and tips forward if a > g b / h, where b is
# the distance from the centre of mass to the front edge of the base
# (measured horizontally) and h is the height of the centre of mass.
g, mu_s, b = 9.81, 0.6, 0.15        # m/s^2, -, m

print("  h (m)   slide at (m/s^2)   tip at (m/s^2)   what happens first")
for h in (0.10, 0.15, 0.20, 0.30, 0.35, 0.45):
    a_slide = mu_s * g
    a_tip = g * b / h
    first = "slides" if a_slide < a_tip else "tips"
    print(f"  {h:4.2f}   {a_slide:10.2f}        {a_tip:10.2f}        {first}")

h_crit = b / mu_s
print(f"tipping governs when h > b / mu_s = {h_crit:.3f} m")

# Braking limit for a shuttle that must not upset any load up to h = 0.45 m,
# with a safety factor of 1.5 on both sliding and tipping:
a_allow = min(mu_s * g, g * b / 0.45) / 1.5
print(f"allowable braking deceleration: {a_allow:.2f} m/s^2 ({a_allow / g:.2f} g)")

  h (m)   slide at (m/s^2)   tip at (m/s^2)   what happens first
  0.10         5.89             14.71        slides
  0.15         5.89              9.81        slides
  0.20         5.89              7.36        slides
  0.30         5.89              4.91        tips
  0.35         5.89              4.20        tips
  0.45         5.89              3.27        tips
tipping governs when h > b / mu_s = 0.250 m
allowable braking deceleration: 2.18 m/s^2 (0.22 g)

An allowable deceleration of about 0.22 g is far below what the shuttle's brakes can deliver. That is typical. The limit on how fast an automated vehicle can stop is often set by its load, not its brakes, and controllers are programmed with separate deceleration limits for normal stops and emergency stops, with the emergency case checked against the worst load the system is allowed to carry.

Rotation about a fixed axis

For a body rotating about a fixed axis through \(O\), the centre of mass moves on a circle, so \((a_G)_t = \alpha r_G\) and \((a_G)_n = \omega^2 r_G\). The moment equation is most conveniently written about the axis itself, where the bearing reactions have no moment:

The force equations then give the bearing reactions. If the centre of mass lies on the axis, as it should for any well-balanced rotor, \(\mathbf{a}_G = 0\) and the bearings carry only the static loads. If it does not, the bearings must supply the force \(m\omega^2 r_G\) to keep the centre of mass on its circle, a force that rotates with the shaft and grows with the square of speed. A rotor with its centre of mass just 0.1 mm off the axis, spinning at 3,000 rev/min, needs a rotating force of about 10 N per kilogram of rotor mass: that is the source of most machine vibration, and the reason rotors are balanced.

Inertia seen through a gear reduction is multiplied by the square of the ratio. A motor rotor with an inertia of only 0.005 kg·m², driving through a 20 to 1 gearbox, looks like \(0.005 \times 20^2 = 2.0 \text{ kg·m}^2\) at the drum, more than the drum itself. Designers of servo systems keep track of this reflected inertia carefully, because a motor whose own inertia dominates the load spends most of its torque accelerating itself.

General plane motion

For general plane motion all three equations are needed, usually together with a kinematic relation between \(a_G\) and \(\alpha\). The commonest case is a wheel, roller or cylinder that may roll, slip or do some of each. The method is to assume rolling without slipping, which gives \(a_G = \alpha r\), solve, and then check whether the friction force required is less than \(\mu_s N\). If it is not, the body slips: replace the rolling relation with \(F = \mu_k N\) and solve again.

The factor of two-thirds is the fraction of the potential energy that goes into translation; the rest goes into spinning. A hoop, with all its mass at the rim, would accelerate at only half the sliding-block value; a solid sphere at five-sevenths. Lesson 11 explains these results with energy, which is often the faster route.