Chapter 5.3, Problem 19E

The ballistic pendulum Historically, in order to maintain quality control over munitions (bullets) produced by an assembly line, the manufacturer would use a ballistic pendulum to determine the muzzle velocity of a gun, that is, the speed of a bullet as it leaves the barrel. Invented in 1742 by the English engineer Benjamin Robins, the ballistic pendulum is simply a plane pendulum consisting of a rod of negligible mass to which a block of wood of mass m_w is attached. The system is set in motion by the impact of a bullet which is moving horizontally at the unknown velocity v_b; at the time of the impact, which we take as t = 0, the combined mass is m_w + m_b, where m_b is the mass of the bullet imbedded in the wood. In (7) of this section, we saw that in the case of small oscillations, the angular displacement θ(t) of a plane pendulum shown in Figure 5.3.3 is given by the linear DE ${\theta }^{″}+(g/l)\theta =0$, where θ > 0 corresponds to motion to the right of vertical. The velocity v_b can be found by measuring the height h of the mass m_w + m_b at the maximum displacement angle θ_max shown in Figure 5.3.11.

Intuitively, the horizontal velocity V of the combined mass (wood plus bullet) after impact is only a fraction of the velocity v_b of the bullet, that is,

$V=\left(\frac{m_b}{m_w+m_b}\right)v_b.$

Now recall, a distance s traveled by a particle moving along a circular path is related to the radius l and central angle θ by the formula s = lθ. By differentiating the last formula with respect to time t, it follows that the angular velocity ω of the mass and its linear velocity v are related by v = lω. Thus the initial angular velocity ω₀ at the time t at which the bullet impacts the wood block is related to V by V = lω₀ or

${\omega }_0=\left(\frac{m_b}{m_w+m_b}\right)\frac{v_b}{l}.$

1. (a) Solve the initial-value problem

$\frac{d^2\theta }{d{t}^2}+\frac{g}{l}\theta =0,\theta (0)=0,{\theta }^{\prime }=(0)={\omega }_0.$

1. (b) Use the result from part (a) to show that

$v_b=\left(\frac{m_w+m_b}{m_b}\right)\sqrt{lg}{\theta }_{\max }.$

1. (c) Use Figure 5.3.11 to express cos θ_max in terms of l and h. Then use the first two terms of the Maclaurin series for cos θ to express θ_max in terms of l and h. Finally, show that v_b is given (approximately) by

$v_b=\left(\frac{m_w+m_b}{m_b}\right)\sqrt{2gh}.$

1. (d) Use the result in part (c) to find v_b and m_b = 5 g, m_w = 1 kg, and h = 6 cm.