Question 17.SP.11: A 0.05-lb bullet B is fired with a horizontal velocity of 15......
A 0.05-lb bullet B is fired with a horizontal velocity of 1500 ft/s into the side of a 20-lb square panel suspended from a hinge at A. Knowing that the panel is initially at rest, determine (a) the angular velocity of the panel immediately after the bullet becomes embedded, (b) the impulsive reaction at A, assuming that the bullet becomes embedded in 0.0006 s.
STRATEGY: Because you have an impact, use the principle of impulse and momentum.
MODELING: Choose your system to be the bullet and the panel, where you model the bullet as a particle and the panel as a rigid body. The impulse–momentum diagram for this system is shown in Fig. 1. Because the time interval Δt = 0.0006 s is very short, you can neglect all non-impulsive forces and consider only the external impulses A _x \Delta t \text { and } A _y \Delta t \text {. }
ANALYSIS:
Principle of Impulse and Momentum.
\text { Syst Momenta }_1+\text { Syst Ext Imp }_{1 \rightarrow 2}=\text { Syst Momenta }_2
+\circlearrowleft \text { moments about } A: \qquad m_B v_B\left(\frac{14}{12}~ft \right)+0=m_P \bar{v}_2\left(\frac{9}{12}~ft \right)+\bar{I}_P \omega_2 (1)
\stackrel{+}{\rightarrow} x \text { components: } \qquad m_B v_B+A_x \Delta t=m_P \bar{v}_2 (2)
+\uparrow y \text { components: } \qquad 0+A_y \Delta t=0 (3)
Note that the weight of the bullet is negligible compared to the weight of the panel, so we did not include it on the right-hand side of Eq. (1). The centroidal mass moment of inertia of the square panel is
\bar{I}_P=\frac{1}{6} m_P b^2=\frac{1}{6} \biggr (\frac{20~lb }{32.2~ft / s ^2} \biggr) \biggr ( \frac{18}{12}~ft \biggr) ^2=0.2329~lb \cdot ft \cdot s ^2
Substituting this value as well as the given data into Eq. (1) and noting that from kinematics, you know
\bar{v}_2=\left(\frac{9}{12}~ft \right) \omega_2
You now have
\biggr( \frac{0.05}{32.2}\biggr) (1500)\left(\frac{14}{12}\right)=0.2329 \omega_2+ \biggr( \frac{20}{32.2}\biggr) \left(\frac{9}{12} \omega_2\right)\left(\frac{9}{12}\right)
\omega_2 = 4.67 rad/s \omega_2 = 4.67 rad/s ↺ ◂
\bar{v}_2=\left(\frac{9}{12}~ft \right) \omega_2=\left(\frac{9}{12}~ft \right)(4.67~rad / s )=3.50~ft / s
Substituting \bar{v}_2 = 3.50 ft/s , Δt = 0.0006 s, and the given data into Eq. (2) gives you
\biggr( \frac{0.05}{32.2}\biggr) (1500)+A_x(0.0006)= \biggr( \frac{20}{32.2} \biggr) (3.50)
A_x=-259~lb \qquad \qquad A _x=259~lb \leftarrow◂
From Eq. (3), you find A_y = 0.
A_y = 0 ◂
REFLECT and THINK: The speed of the bullet is in the range of a modern high-performance rifle. Notice that the reaction at A is over 5000 times the weight of the bullet and over 10 times the weight of the plate.
