Chapter 18.1, Problem 18.32P

Determine the impulse exerted on the plate of Prob. 18.31 during the impact by (a) the obstruction at B, (b) the support at A.

18.31 A square plate of side a and mass m supported by a ball-and-socket joint at A is rotating about the y axis with a constant angular velocity ω = ω₀j when an obstruction is suddenly introduced at B in the xy plane. Assuming the impact at B to be perfectly plastic (e = 0), determine immediately after the impact (a) the angular velocity of the plate, (b) the velocity of its mass center G.

Fig. P18.31

To determine

The impulse exerted $\left(F\Delta t\right)$ on the plate during the impact by obstruction at B.

Answer to Problem 18.32P

The impulse exerted $\left(F\Delta t\right)$ on the plate during the impact by obstruction at B is $\underset{\_}{0.1031ma{\omega }_0}$.

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:

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

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

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.

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 $j^{\prime }$.

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.

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.

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 ${\omega }_{x^{\prime }}$.

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

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)

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)

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

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).

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

Substitute $\frac{1}{8}\sqrt{2}{\omega }_0$ for ${\omega }_{y^{\prime }}$ in Equation (9).

$\begin{array}{l}\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0-\left(aF\Delta t\right)=-\frac{1}{4}m{a}^2\frac{1}{8}\sqrt{2}{\omega }_0\\ aF\Delta t=\frac{1}{24}\sqrt{2}m{a}^2{\omega }_0+\frac{1}{32}\sqrt{2}m{a}^2{\omega }_0\\ aF\Delta t=\frac{7}{96}\sqrt{2}m{a}^2{\omega }_0\\ F\Delta t=0.1031ma{\omega }_0\end{array}$

Therefore, the impulse exerted $\left(F\Delta t\right)$ on the plate during the impact by obstruction at B is $\underset{\_}{0.1031ma{\omega }_0}$.

To determine

Find the impulse exerted $\text{A}\left(\Delta t\right)$ on the plate during the impact by support at B.

Answer to Problem 18.32P

The impulse exerted $\text{A}\left(\Delta t\right)$ on the plate during the impact by support at B is $\underset{\_}{-0.01473ma{\omega }_{0}k}$.

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:

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

Consider the Equation (3).

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

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 ${\omega }_{y^{\prime }}$.

$\begin{array}{c}\overline{v}=\frac{1}{2}a\left(\frac{1}{8}\sqrt{2}{\omega }_0\right)k^{\prime }\\ =\frac{1}{16}\sqrt{2}a{\omega }_0{k}^{\prime }\\ =0.0884a{\omega }_{0}k\end{array}$

The expression for the linear momentum of the system as follows:

$m{\overline{\text{v}}}_0+\text{A}\left(\Delta t\right)+F\left(\Delta t\right)k=m\overline{\text{v}}$

Substitute 0 for ${\overline{\text{v}}}_0$, $\frac{7}{96}\sqrt{2}m{a}^2{\omega }_0$ for $F\left(\Delta t\right)$, and $\frac{1}{16}\sqrt{2}a{\omega }_0{k}^{\prime }$ for $\overline{\text{v}}$.

$\begin{array}{l}m\left(0\right)+\text{A}\left(\Delta t\right)+\frac{7}{96}\sqrt{2}m{a}^2{\omega }_0{k}^{\prime }=\frac{1}{16}\sqrt{2}ma{\omega }_0{k}^{\prime }\\ \text{A}\left(\Delta t\right)=\frac{1}{16}\sqrt{2}ma{\omega }_0{k}^{\prime }-\frac{7}{96}\sqrt{2}m{a}^2{\omega }_0{k}^{\prime }\\ \text{A}\left(\Delta t\right)=-\frac{1}{96}\sqrt{2}ma{\omega }_0{k}^{\prime }\\ \text{A}\left(\Delta t\right)=-0.01473ma{\omega }_{0}k\end{array}$

Therefore, the impulse exerted $\text{A}\left(\Delta t\right)$ on the plate during the impact by support at B is $\underset{\_}{-0.01473ma{\omega }_{0}k}$.