Chapter 18.1, Problem 18.52P

To determine

The kinetic energy (U) lost when plate hits the obstruction at B.

Answer to Problem 18.52P

The kinetic energy (U) lost when plate hits the obstruction at B is $\underset{\_}{\frac{7}{192}m{a}^2{\omega }_0^2}$.

Explanation of Solution

Given information:

The mass of the square plate is m.

The side of a square plate is a.

The angular velocity $\left(\omega \right)$ is ${\omega }_{0}j$

Assume the impact to be perfectly plastic that is $e=0$.

Calculation:

Draw the diagram of the system as in Figure (1).

The length of the diagonal of a square is obtained by multiplying the side with $\sqrt{2}$.

Write the expression for the angular velocity in the $x^{\prime }$ axis $\left({\omega }_{x^{\prime }}\right)$ as follows:

${\omega }_{x^{\prime }}=\frac{\sqrt{2}}{2}{\omega }_0$

Here, ${\omega }_0$ is the initial angular velocity.

Write the expression for the angular velocity in the $y^{\prime }$ axis $\left({\omega }_{y^{\prime }}\right)$ as follows:

${\omega }_{y^{\prime }}=\frac{\sqrt{2}}{2}{\omega }_0$

The unit vectors along the $x^{\prime }$ and $y^{\prime }$ axis are represented by $i^{\prime }$ and $j^{\prime }$.

Write the expression for the initial angular momentum about the mass center ${\left(H_G\right)}_0$ as follows:

${\left(H_G\right)}_0={\overline{I}}_{x^{\prime }}{\omega }_{x^{\prime }}{i}^{\prime }+{\overline{I}}_{y^{\prime }}{\omega }_{y^{\prime }}{j}^{\prime }$ (1)

Here, ${\left(H_G\right)}_0$ is the initial angular momentum about the mass center, ${\overline{I}}_{x^{\prime }}$ is the moment of inertia in the $x^{\prime }$ direction and ${\overline{I}}_{y^{\prime }}$ is the moment of inertia in the $y^{\prime }$ direction.

Write the expression for the moment of inertia in the $x^{\prime }$ direction $\left({\overline{I}}_{x^{\prime }}\right)$ as follows:

${\overline{I}}_{x^{\prime }}=\frac{1}{12}m{a}^2$

Here, m is the mass and a is the side of the square plate.

Due to symmetry, moment of inertia in the $x^{\prime }$ and $y^{\prime }$ axes are the same.

Write the expression for the angular momentum about $z^{\prime }$ axis $\left({\overline{I}}_{z^{\prime }}\right)$ as follows:

${\overline{I}}_{z^{\prime }}=\frac{1}{12}m{a}^2$

Substitute $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{x^{\prime }}$, $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{y^{\prime }}$, $\frac{\sqrt{2}}{2}{\omega }_0$ for ${\omega }_{x^{\prime }}$, and $\frac{\sqrt{2}}{2}{\omega }_0$ for ${\omega }_{y^{\prime }}$ in Equation (1).

$\begin{array}{c}{\left(H_G\right)}_0=\left(\frac{1}{12}m{a}^2\right)\left(\frac{\sqrt{2}}{2}{\omega }_0\right)i^{\prime }+\left(\frac{1}{12}m{a}^2\right)\left(\frac{\sqrt{2}}{2}{\omega }_0\right)j^{\prime }\\ =\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{i}^{\prime }+\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{j}^{\prime }\end{array}$ (2)

Calculate the angular velocity at B $\left(v_B\right)$ using the formula:

$v_B=\omega \times r_{B/A}$

Here, $\omega$ is the angular velocity of the rotating plate and ${\text{r}}_{B/A}$ is the distance of B with respect to A.

Substitute ${\omega }_{x^{\prime }}{i}^{\prime }+{\omega }_{y^{\prime }}{j}^{\prime }+{\omega }_{z^{\prime }}{k}^{\prime }$ for $\omega$ and $-a{j}^{\prime }$ for $r_{B/A}$.

$v_B=\left({\omega }_{x^{\prime }}{i}^{\prime }+{\omega }_{y^{\prime }}{j}^{\prime }+{\omega }_{z^{\prime }}{k}^{\prime }\right)\times \left(-a{j}^{\prime }\right)$

The matrix multiplication for vector product is done.

$v_B=a\left({\omega }_{z^{\prime }}{i}^{\prime }+{\omega }_{x^{\prime }}{k}^{\prime }\right)$

The corner B does not rebound after impact. Therefore, the velocity of B after impact in the $z^{\prime }$ axis ${\left(v_B\right)}_{z^{\prime }}$ is zero. The angular velocity in the $x^{\prime }$ axis (${\omega }_{x^{\prime }}$) is also zero.

Calculate the angular velocity about the mass center $\left(\overline{v}\right)$ using the formula:

$\overline{v}=\text{ω}\times {\text{r}}_{G/A}$

Here, ${\text{r}}_{G/A}$ is the distance of mass center with respect to A.

Substitute ${\omega }_{x^{\prime }}{i}^{\prime }+{\omega }_{y^{\prime }}{j}^{\prime }+{\omega }_{z^{\prime }}{k}^{\prime }$ for $\omega$ and $\frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)$ for ${\text{r}}_{G/A}$.

$\overline{v}=\left({\omega }_{x^{\prime }}{i}^{\prime }+{\omega }_{y^{\prime }}{j}^{\prime }+{\omega }_{z^{\prime }}{k}^{\prime }\right)\times \frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)$

Substitute 0 for ${\omega }_{x^{\prime }}$.

$\overline{v}=\left({\omega }_{y^{\prime }}{j}^{\prime }+{\omega }_{z^{\prime }}{k}^{\prime }\right)\times \frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)$

Write the matrix multiplication for vector product is done.

$\overline{v}=\frac{1}{2}a\left({\omega }_{z^{\prime }}{i}^{\prime }-{\omega }_{z^{\prime }}{j}^{\prime }+{\omega }_{y^{\prime }}{k}^{\prime }\right)$ (3)

Write the expression for the angular momentum about A as follows:

$H_A=H_G+{\text{r}}_{G/A}\times m\overline{\text{v}}$ (4)

Here, $H_A$ is the angular momentum about A and $H_G$ is the angular momentum about the mass center.

Calculate the angular momentum about G using the formula:

$H_G={\overline{I}}_{x^{\prime }}{\omega }_{x^{\prime }}{i}^{\prime }+{\overline{I}}_{y^{\prime }}{\omega }_{y^{\prime }}{j}^{\prime }+{\overline{I}}_{z^{\prime }}{\omega }_{z^{\prime }}{k}^{\prime }$

Substitute 0 for ${\omega }_{x^{\prime }}$, $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{y^{\prime }}$, and $\frac{1}{6}m{a}^2$ for ${\overline{I}}_{z^{\prime }}$.

$H_G=\left(\frac{1}{12}m{a}^2\right){\omega }_{y^{\prime }}{j}^{\prime }+\left(\frac{1}{6}m{a}^2\right){\omega }_{z^{\prime }}{k}^{\prime }$ (5)

Substitute $\left(\frac{1}{12}m{a}^2\right){\omega }_{y^{\prime }}{j}^{\prime }+\left(\frac{1}{6}m{a}^2\right){\omega }_{z^{\prime }}{k}^{\prime }$ for $H_G$, $\frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)$ for ${\text{r}}_{G/A}$, and $\frac{1}{2}a\left({\omega }_{z^{\prime }}{i}^{\prime }-{\omega }_{z^{\prime }}{j}^{\prime }-{\omega }_{y^{\prime }}{k}^{\prime }\right)$ for $\overline{\text{v}}$ in Equation (4).

$H_A=\left[\begin{array}{l}\left(\frac{1}{12}m{a}^2\right){\omega }_{y^{\prime }}{j}^{\prime }+\left(\frac{1}{6}m{a}^2\right){\omega }_{z^{\prime }}{k}^{\prime }+\\ \left[\frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)\times m\frac{1}{2}a\left({\omega }_{z^{\prime }}{i}^{\prime }-{\omega }_{z^{\prime }}{j}^{\prime }-{\omega }_{y^{\prime }}{k}^{\prime }\right)\right]\end{array}\right]$

The matrix multiplication is done for vector product.

$\begin{array}{c}{H}_A=\left[\begin{array}{l}\left(\frac{1}{12}m{a}^2\right){\omega }_{y^{\prime }}{j}^{\prime }+\left(\frac{1}{6}m{a}^2\right){\omega }_{z^{\prime }}{k}^{\prime }+\\ \frac{1}{4}m{a}^2\left(-{\omega }_{y^{\prime }}{i}^{\prime }+{\omega }_{y^{\prime }}{j}^{\prime }+2{\omega }_{z^{\prime }}{k}^{\prime }\right)\end{array}\right]\\ =\left[\begin{array}{l}-\frac{1}{4}m{a}^2{\omega }_{y^{\prime }}{i}^{\prime }+\frac{1}{12}m{a}^2{\omega }_{y^{\prime }}{j}^{\prime }+\frac{1}{4}m{a}^2{\omega }_{y^{\prime }}{j}^{\prime }+\frac{1}{6}m{a}^2{\omega }_{z^{\prime }}{k}^{\prime }+\\ \frac{1}{2}m{a}^2{\omega }_{z^{\prime }}{k}^{\prime }\end{array}\right]\\ =-\frac{1}{4}m{a}^2{\omega }_{y^{\prime }}{i}^{\prime }+\frac{1}{3}m{a}^2{\omega }_{y^{\prime }}{j}^{\prime }+\frac{2}{3}m{a}^2{\omega }_{z^{\prime }}{k}^{\prime }\end{array}$ (6)

The initial velocity of mass center (${\overline{\text{v}}}_0$) is zero.

Calculate the initial momentum about A using the relation:

${\left(H_A\right)}_0={\left(H_G\right)}_0+{\text{r}}_{G/A}\times m{\overline{\text{v}}}_0$

Here, ${\left(H_A\right)}_0$ is the angular momentum of the mass center before impact.

Substitute $\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{i}^{\prime }+\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{j}^{\prime }$ for ${\left(H_G\right)}_0$, $\frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)$ for ${\text{r}}_{G/A}$, and 0 for ${\overline{\text{v}}}_0$.

$\begin{array}{c}{\left(H_A\right)}_0=\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{i}^{\prime }+\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{j}^{\prime }+\frac{1}{2}a\left(-i^{\prime }-j^{\prime }\right)\times 0\\ =\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{i}^{\prime }+\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{j}^{\prime }\end{array}$ (7)

Draw the forces acting on the plate as in Figure (2).

Write the expression for the moment about A as follows:

${\left(H_A\right)}_0+\left(-a{j}^{\prime }\right)\times \left(F\Delta t\right)k^{\prime }=H_A$

The matrix multiplication for vector product is done.

${\left(H_A\right)}_0-\left(aF\Delta t\right)i^{\prime }=H_A$

Substitute Equation (6) and Equation (7).

$\left[\begin{array}{l}\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{i}^{\prime }+\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0{j}^{\prime }\\ -\left(aF\Delta t\right)i^{\prime }\end{array}\right]=\left[\begin{array}{l}-\frac{1}{4}m{a}^2{\omega }_{y^{\prime }}{i}^{\prime }+\frac{1}{3}m{a}^2{\omega }_{y^{\prime }}{j}^{\prime }\\ +\frac{2}{3}m{a}^2{\omega }_{z^{\prime }}{k}^{\prime }\end{array}\right]$ (8)

Equate $i^{\prime }$ components from Equation (8).

$\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0-\left(aF\Delta t\right)=-\frac{1}{4}m{a}^2{\omega }_{y^{\prime }}$ (9)

Equate $j^{\prime }$ components from Equation (8).

$\begin{array}{l}\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0=\frac{1}{3}m{a}^2{\omega }_{y^{\prime }}\\ {\omega }_{y^{\prime }}=\frac{3}{24}\frac{\sqrt{2}m{a}^2{\omega }_0}{m{a}^2}\\ {\omega }_{y^{\prime }}=\frac{1}{8}\sqrt{2}{\omega }_0\end{array}$

Equate $k^{\prime }$ components from Equation (8).

$\begin{array}{l}0=\frac{2}{3}m{a}^2{\omega }_{z^{\prime }}\\ {\omega }_{z^{\prime }}=0\end{array}$

Calculate the velocity along the x, y and z axes $\left(\overline{v}\right)$ using the formula:

$\overline{\text{v}}=\frac{1}{2}a\left({\omega }_{z^{\prime }}{i}^{\prime }-{\omega }_{z^{\prime }}{j}^{\prime }+{\omega }_{y^{\prime }}{k}^{\prime }\right)$

Substitute 0 for ${\omega }_{z^{\prime }}$.

$\begin{array}{c}\overline{\text{v}}=\frac{1}{2}a\left(\left(0\right)i^{\prime }-\left(0\right)j^{\prime }-{\omega }_{y^{\prime }}{k}^{\prime }\right)\\ =\frac{1}{2}a{\omega }_{y^{\prime }}{k}^{\prime }\end{array}$

Substitute $\frac{1}{8}\sqrt{2}{\omega }_0$ for ${\omega }_{y^{\prime }}$.

$\begin{array}{c}\overline{v}=\frac{1}{2}a\left(\frac{1}{8}\sqrt{2}{\omega }_0\right)k^{\prime }\\ =\frac{1}{16}\sqrt{2}a{\omega }_0{k}^{\prime }\end{array}$

Calculate the kinetic energy of the system before impact $\left(T_0\right)$ using the formula:

$T_0=\frac{1}{2}m{\overline{\text{v}}}_0+\frac{1}{2}{\overline{I}}_{x^{\prime }}{\omega }_{{}_{x^{\prime }}}^2+\frac{1}{2}{\overline{I}}_{y^{\prime }}{\omega }_{{}_{y^{\prime }}}^2+\frac{1}{2}{\overline{I}}_{z^{\prime }}{\omega }_{{}_{z^{\prime }}}^2$

Substitute 0 for ${\overline{\text{v}}}_0$, $\frac{1}{8}\sqrt{2}{\omega }_0$ for ${\omega }_{y^{\prime }}$, 0 for ${\omega }_{z^{\prime }}$, 0 for ${\omega }_{x^{\prime }}$, $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{y^{\prime }}$, and $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{x^{\prime }}$.

$\begin{array}{c}{T}_0=0+\frac{1}{2}\left(\frac{1}{12}m{a}^2\right){\left(\frac{\sqrt{2}}{2}{\omega }_0\right)}^2+0+\frac{1}{2}\left(\frac{1}{12}m{a}^2\right){\left(\frac{\sqrt{2}}{2}{\omega }_0\right)}^2+0\\ =\frac{1}{48}m{a}^2{\omega }_0^2+\frac{1}{48}m{a}^2{\omega }_0^2\\ =\frac{1}{24}m{a}^2{\omega }_0^2\end{array}$

Calculate the kinetic energy of the system after impact $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m\overline{\text{v}}+\frac{1}{2}{\overline{I}}_{x^{\prime }}{\omega }_{{}_{x^{\prime }}}^2+\frac{1}{2}{\overline{I}}_{y^{\prime }}{\omega }_{{}_{y^{\prime }}}^2+\frac{1}{2}{\overline{I}}_{z^{\prime }}{\omega }_{{}_{z^{\prime }}}^2$

Substitute $\frac{1}{16}\sqrt{2}a{\omega }_0$ for $\overline{\text{v}}$, $\frac{\sqrt{2}}{8}{\omega }_0$ for ${\omega }_{y^{\prime }}$, 0 for ${\omega }_{z^{\prime }}$, 0 for ${\omega }_{x^{\prime }}$, $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{y^{\prime }}$, and $\frac{1}{12}m{a}^2$ for ${\overline{I}}_{x^{\prime }}$.

$\begin{array}{c}{T}_0=\frac{1}{2}m{\left(\frac{1}{16}\sqrt{2}a{\omega }_0\right)}^2+0+\frac{1}{2}\left(\frac{1}{12}m{a}^2\right){\left(\frac{\sqrt{2}}{8}{\omega }_0\right)}^2+0\\ =m{a}^2{\omega }_0^2\left(\frac{1}{256}+\frac{1}{768}\right)\\ =m{a}^2{\omega }_0^2\left(\frac{4}{768}\right)\\ =\frac{1}{192}m{a}^2{\omega }_0^2\end{array}$

Calculate the loss in kinetic energy (U) using the formula:

$U=T_0-T_1$

Substitute $\frac{1}{24}m{a}^2{\omega }_0^2$ for $T_0$ and $\frac{1}{192}m{a}^2{\omega }_0^2$ for $T_1$.

$\begin{array}{c}U=\frac{1}{24}m{a}^2{\omega }_0^2-\frac{1}{192}m{a}^2{\omega }_0^2\\ =\frac{8-1}{192}m{a}^2{\omega }_0^2\\ =\frac{7}{192}m{a}^2{\omega }_0^2\end{array}$

Thus, the kinetic energy (U) lost when plate hits the obstruction at B is $\underset{\_}{\frac{7}{192}m{a}^2{\omega }_0^2}$.