Plane Motion of Rigid Bodies: Work and Energy Method

Kinetic energy of a rigid body in plane motion

Let the body have mass $m$, angular velocity $\omega$, and let the mass centre G have velocity $\mathbf{v}_G$. A particle of mass $dm$ at position $\boldsymbol\rho$ relative to G has velocity $\mathbf{v}_G + \boldsymbol\omega\times\boldsymbol\rho$ (velocity of G plus velocity relative to G, magnitude $\omega\rho$).

$$T = \int \tfrac12 |\mathbf{v}_G + \boldsymbol\omega\times\boldsymbol\rho|^2\,dm = \tfrac12 v_G^2\int dm + \mathbf{v}_G\cdot\left(\boldsymbol\omega\times\int\boldsymbol\rho\,dm\right) + \tfrac12\omega^2\int\rho^2\,dm$$

Since G is the mass centre, $\int\boldsymbol\rho\,dm = 0$, so the middle term vanishes, and $\int\rho^2\,dm = I_G$:

$$T = \tfrac12 m v_G^2 + \tfrac12 I_G\,\omega^2$$

Special cases: pure translation $T = \tfrac12mv^2$; rotation about a fixed point O, $T = \tfrac12 I_O\omega^2$; rolling without slipping, $T = \tfrac12 I_C\omega^2$ about the contact point.

Principle of work and energy for a rigid body

$$T_1 + U_{1\to2} = T_2$$

where $U_{1\to2}$ is the work of all external forces and couples (and internal ones that do work, such as springs or friction between connected parts) between positions 1 and 2. For a system of connected bodies, add the kinetic energies of all bodies.

Work of forces and couples

• Force $\mathbf{F}$: $U = \int \mathbf{F}\cdot d\mathbf{r}$ at its point of application; weight: $U = -W\Delta h$; spring: $U = -\tfrac12 k(x_2^2 - x_1^2)$.
• Couple $M$: $U = \int M\,d\theta$, which is $M\Delta\theta$ for constant $M$.

Forces that do no work

• Reactions at smooth fixed pins and smooth surfaces (perpendicular to motion).
• The friction force on a body rolling without slipping, because its point of application (the contact point) is instantaneously at rest, so no energy is dissipated. The normal reaction on a rolling body also does no work.
• Internal forces between particles of a rigid body (equal, opposite, no relative displacement).

If the body slips, the friction force does work $-\mu N\,s$ (energy lost as heat).

Given: $m = 15$ kg, $r = 0.25$ m, $k = 400$ N/m, initial spring extension $x = 0.3$ m (released from rest). Rolling without slipping, so $\omega = v/r$.

  |/\/\/\/\/\---(o)-->       cylinder rolls, spring
  |  k = 400            unstretched at the mid position

Moment of inertia

$$I_G = \tfrac12 mr^2 = \tfrac12(15)(0.25)^2 = 0.4688\ \text{kg·m}^2$$

Work done

Friction at the rolling contact does no work and the normal reaction and weight do no work (the motion is horizontal). Only the spring does work, as it goes from extension 0.3 m to zero:

$$U_{1\to2} = \tfrac12 k x^2 = \tfrac12(400)(0.3)^2 = 18.0\ \text{J}$$

Kinetic energy

Position 1: at rest, $T_1 = 0$.

Position 2 (spring unstretched):

$$T_2 = \tfrac12 m v^2 + \tfrac12 I_G\omega^2 = \tfrac12 mv^2 + \tfrac12\left(\tfrac12 mr^2\right)\frac{v^2}{r^2} = \tfrac34 mv^2$$

Work-energy equation

$$0 + 18.0 = \tfrac34(15)v^2 = 11.25\,v^2$$ $$v = \sqrt{\frac{18.0}{11.25}} = 1.265\ \text{m/s} \qquad \omega = \frac{v}{r} = \frac{1.265}{0.25} = 5.06\ \text{rad/s}$$

Check of the energy split: translational $\tfrac12mv^2 = 12.0$ J and rotational $\tfrac14 mv^2 = 6.0$ J, total $18.0$ J.

Answer: v = 1.26 m/s; ω = 5.06 rad/s.

Given: drum $M = 20$ kg, $r = 0.30$ m, $k = 0.25$ m; block $m = 10$ kg; fall $h = 2$ m from rest.

      ___
     /   \  drum (fixed axle)
     \___/|
          | rope
         [m]  falls h = 2 m

Drum moment of inertia

$$I = Mk^2 = 20(0.25)^2 = 1.25\ \text{kg·m}^2 \qquad \frac{I}{r^2} = 13.889\ \text{kg}$$

Work-energy method

Rope tension does no net work on the system (it acts on both the block and the drum with opposite signs); only the weight of the block does work:

$$U = mgh = 10(9.81)(2) = 196.2\ \text{J}$$

Final kinetic energy, with $\omega = v/r$:

$$T_2 = \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 v^2\left(m + \frac{I}{r^2}\right) = \tfrac12 v^2(10 + 13.889)$$ $$0 + 196.2 = \tfrac12 v^2(23.889) \;\Rightarrow\; v = 4.053\ \text{m/s}$$

Acceleration of the block

For constant acceleration from rest, $v^2 = 2ah$:

$$a = \frac{v^2}{2h} = \frac{(4.053)^2}{4} = 4.107\ \text{m/s}^2$$

Check: $a = \dfrac{mg}{m + I/r^2} = \dfrac{98.1}{23.889} = 4.107$ m/s².

Angular acceleration and tension

$$\alpha = \frac{a}{r} = \frac{4.107}{0.3} = 13.69\ \text{rad/s}^2 \qquad T = m(g - a) = 10(9.81 - 4.107) = 57.03\ \text{N}$$

Answer: v = 4.05 m/s after 2 m; a = 4.11 m/s²; α = 13.69 rad/s²; T = 57.0 N.