Chapter 17.3, Problem 17.115P

The uniform rectangular block shown is moving along a frictionless surface with a velocity ${\bar{v}}_1$ when it strikes a small obstruction at B. Assuming that the impact between corner A and obstruction B is perfectly plastic, determine the magnitude of the velocity ${\bar{v}}_1$ for which the maximum angle θ through which the block will rotate will be 30°.

Fig. P17.115

To determine

Find the magnitude of the velocity ${\overline{v}}_1$ for which the maximum angle q through which the block will rotate $30°$.

Answer to Problem 17.115P

The magnitude of the velocity ${\overline{\text{v}}}_1$ for which the maximum angle $\theta$ through which the block will rotate $30°$ is $\underset{\_}{2.38\text{m/s}}$.

Explanation of Solution

Given information:

The length $\left(a\right)$ of rectangular block is 200 mm.

The width $\left(b\right)$ of rectangular block is 100 mm.

Calculation:

Write the equation of centroidal moment of inertia $\left(\overline{I}\right)$ of the rectangular block using the equation:

$\overline{I}=\frac{1}{12}m\left(a^2+b^2\right)$

Find the one half of the diagonal distance $\left(d\right)$ of the rectangular block.

$d=\frac{1}{2}\sqrt{a^2+b^2}$

Substitute 200 mm for a and 100 mm for b.

$\begin{array}{c}d=\frac{1}{2}\sqrt{{\left(200\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2+{\left(100\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ =\frac{1}{2}\sqrt{{\left(0.200\right)}^2+{\left(0.100\right)}^2}\\ =0.1118\text{m}\end{array}$

Before impact, let ${\overline{\text{v}}}_1$ be the velocity of the rectangular block, and ${\omega }_1$ be the angular velocity of the rectangular block $\left({\omega }_1=0\right)$. Therefore,

${\overline{\text{v}}}_1=v_1\to$

Here, magnitude of velocity of the rectangular block is $v_1$.

After impact, the rectangular block rotates about small obstruction at B. Therefore,

Let ${\omega }_2$ be the angular velocity of the rectangular block after impact.

Find the equation of velocity $\left(v_2\right)$ of the rectangular block after impact.

$v_2=d{\omega }_2$

Consider the principle of impulse and momentum.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the impulse and momentum diagram of the rectangular block as shown in Figure (1).

Take moment about B (positive sign in clockwise direction).

$\begin{array}{l}\frac{m{v}_{1}b}{2}+0=\overline{I}{\omega }_2+md{v}_2\\ \frac{m{v}_{1}b}{2}=\overline{I}{\omega }_2+md{v}_2\end{array}$

Substitute $\frac{1}{12}m\left(a^2+b^2\right)$ for $\overline{I}$, and $d{\omega }_2$ for $v_2$.

$\begin{array}{l}\frac{m{v}_{1}b}{2}=\left(\frac{1}{12}m\left(a^2+b^2\right)\right){\omega }_2+md\left(d{\omega }_2\right)\\ \frac{1}{2}m{v}_{1}b=\left(\frac{1}{12}m\left(a^2+b^2\right)\right){\omega }_2+m{d}^2{\omega }_2\end{array}$

Substitute $\frac{1}{2}\sqrt{a^2+b^2}$ for d.

$\begin{array}{l}\frac{1}{2}m{v}_{1}b=\left(\frac{1}{12}m\left(a^2+b^2\right)\right){\omega }_2+m{\left(\frac{1}{2}\sqrt{a^2+b^2}\right)}^2{\omega }_2\\ \frac{1}{2}m{v}_{1}b=\frac{1}{12}m\left(a^2+b^2\right){\omega }_2+\frac{1}{4}m\left(a^2+b^2\right){\omega }_2\\ \frac{1}{2}m{v}_{1}b=\frac{1}{3}m\left(a^2+b^2\right){\omega }_2\\ {\omega }_2=\frac{3m{v}_{1}b}{2m\left(a^2+b^2\right)}\\ {\omega }_2=\frac{3v_{1}b}{2\left(a^2+b^2\right)}\end{array}$ (1)

Sketch the free body diagram of the rectangular block after impact (position 2) and final position (position 3) as shown in Figure (2).

Write the equation of velocity $\left({\overline{v}}_2\right)$ of the rectangular block immediately after the impact:

${\overline{v}}_2=d{\omega }_2$

Refer Figure (2),

Find the angle $\left(\beta \right)$ when the rectangular block is at position 3.

$\beta ={\tan }^{-1}\frac{b}{a}$

Substitute 200 mm for a and 100 mm for b.

$\begin{array}{c}\beta ={\tan }^{-1}\frac{100\text{mm}}{200\text{mm}}\\ ={\tan }^{-1}0.5\\ =26.565°\end{array}$

Find the distance (h).

$h=d\sin \left(\beta +30°\right)$

Substitute 0.1118 m for d and $26.565°$ for $\beta$.

$\begin{array}{c}h=\left(0.1118\text{m}\right)\sin \left(26.565°+30°\right)\\ =\left(0.1118\text{m}\right)\sin \left(56.565°\right)\\ =0.0933\text{m}\end{array}$

At position 3, the angular velocity $\left({\omega }_3\right)$ will be zero and the velocity of the rectangular block $\left({\overline{v}}_3\right)$ will also be zero.

Write the equation of the potential energy $\left(V_2\right)$ of the system after the impact.

$V_2=mg\frac{b}{2}$

Write the equation of the kinetic energy $\left(T_2\right)$ of the system after the impact.

$T_2=\frac{1}{2}\overline{I}{\omega }_2{}^2+\frac{1}{2}m{\overline{v}}_2{}^2$

Substitute $\frac{1}{12}m\left(a^2+b^2\right)$ for $\overline{I}$ and $d{\omega }_2$ for $v_2$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(\frac{1}{12}m\left(a^2+b^2\right)\right){\omega }_2{}^2+\frac{1}{2}m{\left(d{\omega }_2\right)}^2\\ =\frac{1}{24}m\left(a^2+b^2\right){\omega }_2{}^2+\frac{1}{2}m{d}^2{\omega }_2{}^2\end{array}$

Substitute $\frac{1}{2}\sqrt{a^2+b^2}$ for d.

$\begin{array}{c}{T}_2=\frac{1}{24}m\left(a^2+b^2\right){\omega }_2{}^2+\frac{1}{2}m{\left(\frac{1}{2}\sqrt{a^2+b^2}\right)}^2{\omega }_2{}^2\\ =\frac{1}{24}m\left(a^2+b^2\right){\omega }_2{}^2+\frac{1}{8}m\left(a^2+b^2\right){\omega }_2{}^2\\ =\frac{4}{24}m\left(a^2+b^2\right){\omega }_2{}^2\\ =\frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2\end{array}$

Write the equation of the potential energy $\left(V_3\right)$ of the system at final position:

$V_3=mgh$

The kinetic energy of the system at final position $\left(T_3\right)$ will be zero.

Consider the conservation of energy.

$T_2+V_2=T_3+V_3$

Substitute $\frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2$ for $T_2$, $mg\frac{b}{2}$ for $V_2$, 0 for $T_3$, and $mgh$ for $V_3$.

$\begin{array}{l}\frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2+\left(mg\frac{b}{2}\right)=0+\left(mgh\right)\\ \frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2+\frac{1}{2}mgb=mgh\\ \frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2=mgh-\frac{1}{2}mgb\\ \frac{1}{6}m\left(a^2+b^2\right){\omega }_2{}^2=\frac{mg}{2}\left(2h-b\right)\end{array}$

$\begin{array}{l}{\omega }_2{}^2=\frac{6mg}{2m\left(a^2+b^2\right)}\left(2h-b\right)\\ {\omega }_2{}^2=\frac{3g\left(2h-b\right)}{\left(a^2+b^2\right)}\end{array}$

Substitute 200 mm for a, 100 mm for b, 0.1118 m for d, $9.81{\text{m}/\text{s}}^2$ for g, and 0.0933 m for h.

$\begin{array}{l}{\omega }_2{}^2=\frac{3\left(9.81\right)\left[2\left(0.0933\right)-\left(100\right)\right]}{{\left(200\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2+{\left(100\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ {\omega }_2{}^2=\frac{3\left(9.81\right)\left[2\left(0.0933\right)-\left(100\right)\right]}{{\left(0.200\right)}^2+{\left(0.100\right)}^2}\\ {\omega }_2{}^2=\frac{2.5486}{0.05}{\text{rad}}^2/{\text{s}}^2\\ {\omega }_2=7.1396\text{rad}/\text{s}\end{array}$

Find the the magnitude of the velocity ${\overline{v}}_1$ for which the maximum angle q through which the block will rotate $30°$ using Equation (1).

${\omega }_2=\frac{3v_{1}b}{2\left(a^2+b^2\right)}$

Substitute 200 mm for a, 100 mm for b, and $7.1396\text{rad}/\text{s}$ for ${\omega }_2$.

$\begin{array}{l}7.1396=\frac{3v_1\left(100\times \frac{1\text{m}}{1,000\text{mm}}\right)}{2{\left(200\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2+{\left(100\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ 7.1396=\frac{3v_1\left(0.100\right)}{2\left[{\left(0.200\right)}^2+{\left(0.100\right)}^2\right]}\\ 7.1396=3v_1\\ v_1=\frac{7.1396}{3}\\ v_1=2.38\text{m}/\text{s}\end{array}$

Thus, the magnitude of the velocity ${\overline{\text{v}}}_1$ for which the maximum angle $\theta$ through which the block will rotate $30°$ is $\underset{\_}{2.38\text{m/s}}$.