Chapter 17.3, Problem 17.120P

For the beam of Prob. 17.119, determine the velocity of the 1-oz bullet for which the maximum angle of rotation of the beam will be 90°.

To determine

The velocity of the bullet for which the maximum angle of rotation of the beam will be ${90}^0$

Answer to Problem 17.120P

Velocity of a bullet for ${90}^0$ of the rotation angle of the beam will be $1887ft/sec$.

Explanation of Solution

Given information:

$\theta m=$${90}^0$

Mass of bullet w’ $=1oz=1ounce=0.06251b.$

Mass of beam $w=18lb.$

Velocity of beam $v_0=750mi/h=1100ft/sec$

Concept used:

Moment of inertia $=I=1/12m{l}^2$

Low of conservation of energY

Calculation:

Mass ratio = $\beta =w’/w=0.0625/18=0.00347$

We can say that $w’=\beta w$ or

$M’=\beta m$

Since $\beta$ is so small mass of the bullet is ignored while calculating the motion of beam after impact.

As per kinetics,

Linear component, $M’=-\beta m{v}_0+0=m{v}_{2}m$

$\begin{array}{l}{v}_2=\beta v_0+0=m{v}_2\hfill \\ v_2=\beta v_0\hfill \end{array}$

Taking moment at B,

$\begin{array}{l}0+0=Iw-m{v}_{2}L/2\hfill \\ Iw=m{v}_{2}L/2\hfill \\ W=m{v}_{2}L/2I\hfill \\ =12/2xm{v}_{2}L/m{L}^2\hfill \\ =6v_2/L\hfill \\ W=6\beta v_0/L\hfill \end{array}$

Motion after impact,

$\begin{array}{l}{v}_2=-mgL/2\hfill \\ T_2=1/2I{w}_2{}^2+1/2m{v}_2{}^2\hfill \\ =1/2\left(1/12m{L}^2\right){(6\beta v_0/L)}^2+1/2m{(\beta v_0)}^2\hfill \\ =m{L}^2/24(36\left({\beta }^2{v}_0{}^2/L^2\right)+m/2{\beta }^2{v}_0{}^2\hfill \\ =3m{\beta }^2{v}_0{}^2/2+m{\beta }^2{v}_0{}^2/2\hfill \\ =4/2m{\beta }^2{v}_0{}^2\hfill \\ T_2=2m{\beta }^2{v}_0{}^2\hfill \end{array}$

New position:

$W=0andQ=Q_{max}$

According to figure, $v_3=-mgL/2cos{\theta }_m$

$T_3=1/2m{v}_3{}^2$

Linear components,

$\begin{array}{l}M{v}_2+0=m{v}_3\hfill \\ V_3=v_2=\beta v_0\hfill \end{array}$

Using low of conservation of energy,

$T_3+V_3=T_3+V_3$

$2m{\beta }^2{v}_0{}^2-mgL/2=1/2m\left(\beta v_0{}^2\right)-mgL/2cos{\theta }_m$

$2m\left(\beta v_0{}^2\right)-mgL/2=1/2mgL/2cos{\theta }_m=m\left(\beta v_0{}^2\right)/2$

$2m\left(\beta v_0{}^2\right)-mgL/2\left(1-cos{\theta }_m\right)=m{\left(\beta v_0\right)}^2/2$

$m\left[2\left(\beta v_0{}^2\right)-gL/2\left(1-cos{\theta }_m\right)\right]=m\left(\beta v_0{}^2\right)/2$

$2m/2\left[2\left(\beta v_0{}^2\right)-gL/2\left(1-cos{\theta }_m\right)\right]=\left(\beta v_0{}^2\right)$

$2m\left(\beta v_0{}^2\right)-mgL/2\left(1-cos{\theta }_m\right)={(\beta v_0)}^2{}^{}$

$2m\left(\beta v_0{}^2\right)-{\left(\beta v_0\right)}^2=mgL/2\left(1-cos{\theta }_m\right)$

${\left(\beta v_0\right)}^2\left(2m-1\right)=mgL/2\left(1-cos{\theta }_m\right)$

${\left(\beta v_0\right)}^2=mgL/2\left(1-cos{\theta }_m\right)/4m-2$

${\left(\beta v_0\right)}^2=1/3gL\left(1-cos{\theta }_m\right)\beta v_0$

$=1/3gL/2\left(1-cos{\theta }_m\right)$

$=1/3\left(32.2\right)\left(4\right)\left(1-cos90\right)$

$\begin{array}{l}\beta v_0=6.55ft/sec\hfill \\ v_0=6.55/0.00347\hfill \\ v_0=1887.1ft/sec\hfill \end{array}$

Conclusion:

We can find that bullet velocity for a maximum rotation angle of beam ${90}^0$ is $1887ft/sec$.