A gun of mass M fires a shell S of mass m with a muzzle velocity vSG = Vo relative to the gun barrel G, at an elevation angle α in the ground frame Φ = {F ; Ik} in Fig. 8.1. The gun carriage C is mounted on a greased horizontal track. (a) Find the instantaneous recoil velocity VGF of the gun

Question

A gun of mass M fires a shell S of mass m with a muzzle velocity [latex]\mathbf{v}_{S G}=\mathbf{v}_0[/latex] relative to the gun barrel G, at an elevation angle α in the ground frame [latex]\Phi=\left\{F ; \mathbf{I}_k\right\}[/latex] in Fig. 8.1. The gun carriage C is mounted on a greased horizontal track. (a) Find the instantaneous recoil velocity [latex]\mathbf{v}_{G F}[/latex] of the gun (i.e. the center of mass of the gun assembly ). (b) Compare the magnitude v of the instantaneous, absolute muzzle velocity [latex]\mathbf{V}_{S F}[/latex] of the shell in [latex]\Phi[/latex] with its relative value [latex]v_0[/latex]. (c) What is the instantaneous impulsive reaction exerted by the track on the gun carriage?

Accepted Answer

Solution of (a). The gun and shell are modeled as center of mass objects—“a system of two particles .” Then the instantaneous, absolute muzzle velocity of the shell S relative to the ground frame [latex]\Phi=\left\{F ; \mathbf{I}_k ight\} ext { is } \mathbf{v}_{S F}=\mathbf{v}_{S G} + \mathbf{v}_{G F} ;[/latex] that is,

[latex]\mathbf{v}_{S F}=\mathbf{v}_0 + \mathbf{v}_{G F}=\left(v_0 \cos \alpha - V ight) \mathbf{i} + v_0 \sin \alpha \mathbf{j}[/latex], (8.18a)

wherein the instantaneous recoil velocity of the gun is [latex]\mathbf{v}_{G F}=-V \mathbf{i}[/latex]. To find [latex]\mathbf{v}_{G F}[/latex], we observe that the only external forces that act on the system at the impulsive instant are vertical forces—the total weight of the system, the equipollent static track reaction force on the carriage, and the additional impulsive vertical reaction force R exerted on the gun carriage by the lubricated track . Therefore, in accordance with (8.6), the component [latex]\mathbf{p}^* \cdot \mathbf{i}[/latex] of the linear momentum of the center of mass, hence the system, is conserved. Initially, [latex]\mathbf{p}^*=\mathbf{0}[/latex]; hence, after the impulse, the component [latex]\mathbf{p}^* \cdot \mathbf{i}[/latex] of the instantaneous momentum of the system must vanish in [latex]\Phi[/latex]:

[latex]\mathbf{p}^* \cdot \mathbf{i}=\left(M \mathbf{v}_{G F} + m \mathbf{v}_{S F} ight) \cdot \mathbf{i}=\left(-M V \mathbf{i} + m \mathbf{v}_{S F} ight) \cdot \mathbf{i}=0[/latex] (8.18b)

Equations (8.18a) and (8.18b) yield [latex]-M V + m\left(v_0 \cos \alpha - V ight)=0[/latex]; and hence the instantaneous recoil velocity of the gun is

[latex]\mathbf{v}_{G F}=-V \mathbf{i}=-\frac{m}{m + M} v_0 \cos \alpha \mathbf{i} .[/latex] (8.18c)

After the impulsive instant , additional forces exerted by a recoil spring and viscous damper, not shown in the diagram , retard the subsequent motion of the gun and restore it to its firing station. We shall not explore this motion.

Solution of (b). The instantaneous, absolute muzzle velocity [latex]\mathbf{v}_{S F}[/latex] of the shell follows from (8.18a):

[latex]\mathbf{v}_{S F}=v_0\left(\frac{M}{m + M} \cos \alpha \mathbf{i} + \sin \alpha \mathbf{j} ight)[/latex]; (8.18d)

and hence the absolute muzzle speed v is related to the relative speed [latex]v_0[/latex] by

[latex]v=v_0\left(1 - \frac{m(m + 2 M)}{(m + M)^2} \cos ^2 \alpha ight)^{1 / 2}[/latex]. (8.l8e)

Inasmuch as [latex]m \ll M[/latex], to the first order in [latex]\mu \equiv m / M, v=v_0\left(1-\mu \cos ^2 \alpha ight)

Solution of (c). The instantaneous, external normal reaction impulse [latex]\mathscr{T}_R^*[/latex] exerted by the smooth track on the system is obtained from (8. 17):

[latex]\Delta \mathbf{p}^*=\mathscr{\mathscr { T }}\left(t ; t_0 ight)[/latex] (8. 17)

[latex]\mathscr{T}_R^* \equiv \lim _{t ightarrow t_0} \int_{t_0}^t \mathbf{R} d t=\Delta \mathbf{p}^*=M \mathbf{v}_{G F} + m \mathbf{v}_{S F} .[/latex] (8.18f)

Since [latex]\mathbf{R} \cdot \mathbf{i}=0[/latex], (8.18f) requires . [latex]\Delta \mathbf{p}^* \cdot \mathbf{i}=0[/latex], which is the same as (8.18b); and hence, by (8.l8c) and (8.18d), the external impulsive reaction of the track on the gun is

[latex]\mathscr{T}_R^*=\left(\Delta \mathbf{p}^* \cdot \mathbf{j} ight) \mathbf{j}=m v_0 \sin \alpha \mathbf{j} .[/latex] (8.18g)

Question 8.1: A gun of mass M fires a shell S of mass m with a muzzle velo...

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