The Ballistic Pendulum Historically, to maintain quality

Chapter 5, Problem 19E

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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 \(m_{b}\) 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 \(\theta(t)\) of a plane pendulum shown in Figure 5.3.3 is given by the linear DE \(\theta^{\prime \prime}+(g / l) \theta=0\), where \(\theta>0\) 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 \(\theta_{\max }\) shown in Figure 5.3.10. 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 \(\boldsymbol{\theta}\) by the formula \(s=l \theta\). By differentiating the last formula with respect to time t, it follows that the angular velocity \(\omega\) of the mass and its linear velocity v are related by \(v=l \omega\). Thus the initial angular velocity \(\omega_{0}\) at the time t at which the bullet impacts the wood block is related to V by \(V=l \omega_{0}\) or

\(\omega_{0}=\left(\frac{m_{b}}{m_{w}+m_{b}}\right) \frac{v_{b}}{l}\).

(a) Solve the initial-value problem

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

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

\(v_{b}=\left(\frac{m_{w}+m_{b}}{m_{b}}\right) \sqrt{l g} \theta_{\max }\).

(c) Use Figure 5.3.10 to express cos \(\theta_{\max }\) in terms of l and h. Then use the first two terms of the Maclaurin series for cos \(\theta\) to express \(\theta_{\max }\) in terms of l and h. Finally, show that \(v_{b}\) is given (approximately) by

\(v_{b}=\left(\frac{m_{w}+m_{b}}{m_{b}}\right) \sqrt{2 g h}\).

(d) Use the result in part (c) to find \(v_{b}\) and \(m_{b}=5 \mathrm{~g}, m_{w}=1 \mathrm{~kg}\), and \(h=6 \mathrm{~cm}\).

Text Transcription:

m_w

v_b

m_w+m_b

m_b

theta(t)

theta^prime\prime+(g/l)theta=0

theta>0

theta_max

theta

omega

v=lomega

omega_0

V=lomega_0

omega_0=(fracm_bm_w+m_b)fracv_b)

fracd^2thetadt^2+fracgltheta=0,theta(0)=0,theta^prime(0)=omega_0

theta_max