Plane Motion of Rigid Bodies: Impulse and Momentum Method

Impulse and momentum for a rigid body in plane motion

The linear and angular momenta of the body are $m\mathbf{v}_G$ and $I_G\omega$ (angular momentum about G). The principle states that the initial momenta plus the impulses of the external forces equal the final momenta:

$$m\mathbf{v}_{G1} + \sum\int\mathbf{F}\,dt = m\mathbf{v}_{G2}$$ $$I_G\omega_1 + \sum\int M_G\,dt = I_G\omega_2$$

Angular impulse can be taken about any fixed point O, with the moment of the linear momentum included: $H_O = I_G\omega + m v_G d$ (d is the perpendicular distance from O to the line of $\mathbf{v}_G$). For fixed-axis rotation, $H_O = I_O\omega$.

Conservation

• Linear momentum is conserved in a direction if the external linear impulse in that direction is zero.
• Angular momentum about a point (or axis) is conserved if the angular impulse of external forces about it is zero, for example for forces whose lines pass through the point, or for an isolated rotating system (a diver, an ice skater, a man walking on a platform).

Eccentric impact

Eccentric impact occurs when the line of impact does not pass through the mass centre of one or both colliding bodies; the impulse then produces a change in both linear and angular velocity. The usual case is a particle (or body) striking a rigid body which is free to rotate about a pin.

The following equations are used (smooth contact, impulse $P$ along the line of impact):

1. Linear impulse-momentum for each body along the line of impact.
2. Angular impulse-momentum for the rotating body, about its pin (or G), with the impulse arm $d$: $P\,d = I\,(\omega_2 - \omega_1)$. For two bodies, angular momentum of the system about the pin is conserved if the pin impulse has no moment about the pin.
3. Restitution: $e = \dfrac{(v_2' - v_1')_n}{(v_1 - v_2)_n}$, the ratio of the separation speed to the approach speed of the contact points, measured along the line of impact.

The impact reaction at the pin is an impulsive force unless the impact is at the centre of percussion, where the pin impulse is zero.

Given: platform $M = 40$ kg, $R = 1.5$ m (disc), man $m = 60$ kg, initial speed $N_1 = 20$ rpm. No external torque acts about the vertical axis (the platform rotates freely), so angular momentum is conserved.

Moments of inertia

• Platform: $I_p = \tfrac12 MR^2 = \tfrac12(40)(1.5)^2 = 45.0$ kg·m²
• Man at the centre: contributes $\approx 0$. Initial $I_1 = 45.0$ kg·m².
• Man at the rim: $mR^2 = 60(1.5)^2 = 135$ kg·m². Final $I_2 = 45 + 135 = 180.0$ kg·m².

Final angular speed

$$I_1\omega_1 = I_2\omega_2 \;\Rightarrow\; N_2 = N_1\frac{I_1}{I_2} = 20\times\frac{45}{180} = 5.00\ \text{rpm}$$

In rad/s: $\omega_1 = 2.094$ and $\omega_2 = 0.524$ rad/s.

Change in kinetic energy

$$T_1 = \tfrac12 I_1\omega_1^2 = \tfrac12(45)(2.094)^2 = 98.7\ \text{J}$$ $$T_2 = \tfrac12 I_2\omega_2^2 = \tfrac12(180)(0.524)^2 = 24.7\ \text{J}$$ $$\Delta T = T_2 - T_1 = -74.0\ \text{J}$$

Reason

Angular momentum is conserved, but kinetic energy is not, because the man does internal work (muscular effort and friction between his feet and the platform) while he moves radially outward. The system's energy decreases by 74.0 J; that energy is dissipated, mainly as heat in the man's body and at the feet. Since $T = \tfrac12 I\omega^2 = H^2/2I$, an increase in $I$ at constant $H$ always reduces $T$.

Answer: New speed = 5.0 rpm; kinetic energy decreases by 74.0 J.

Given: rod $M = 3$ kg, $L = 1.2$ m, pivot O (smooth); ball $m = 0.5$ kg, $u = 10$ m/s, strikes at the free end (distance $L$ from O); $e = 0.6$.

     O (pin)
     |
     |  rod M
     |
     |
  -->o  ball m, u = 10 m/s     (impact at the lower end)

Moment of inertia of the rod about O

$$I_O = \frac{ML^2}{3} = \frac{3(1.2)^2}{3} = 1.44\ \text{kg·m}^2$$

Equations

Let $v$ be the velocity of the ball after impact (positive in the original direction) and $\omega$ the angular velocity of the rod after impact. The pin reaction is impulsive, but it passes through O, so angular momentum about O is conserved for the system of ball plus rod:

$$m u L = m v L + I_O\omega$$ $$0.5(10)(1.2) = 0.5\,v\,(1.2) + 1.44\,\omega \;\Rightarrow\; 6 = 0.6v + 1.44\omega \quad (1)$$

Restitution (relative speed of separation at the contact point, rod tip speed $= \omega L$):

$$\omega L - v = e\,u = 0.6(10) = 6 \;\Rightarrow\; v = 1.2\omega - 6 \quad (2)$$

Substituting (2) in (1): $6 = 0.6(1.2\omega - 6) + 1.44\omega = 0.72\omega - 3.6 + 1.44\omega$

$$9.6 = 2.16\,\omega \;\Rightarrow\; \omega = 4.444\ \text{rad/s}$$ $$v = 1.2(4.444) - 6 = -0.667\ \text{m/s}$$

The negative sign means the ball rebounds (moves back) at 0.667 m/s. The rod tip moves at $\omega L = 5.333$ m/s.

Kinetic energy lost

$$T_1 = \tfrac12(0.5)(10)^2 = 25.00\ \text{J} \qquad T_2 = \tfrac12(0.5)(0.667)^2 + \tfrac12(1.44)(4.444)^2 = 14.33\ \text{J}$$ $$\Delta T = 10.67\ \text{J}$$

Answer: ω = 4.44 rad/s; ball rebounds at 0.67 m/s; energy lost = 10.67 J.