Chapter 18.1, Problem 18.51P

To determine

The kinetic energy lost of the plate when edge C of the plate hits the obstruction.

Answer to Problem 18.51P

The energy loss of circular disc after impact is $\frac{m{\overline{v}}_0{}^2}{10}$.

Explanation of Solution

Given information:

Mass of circular plate is $m$, falling velocity of the plate is ${\overline{v}}_0$, impact is perfectly plastic $\left(e=0\right)$.

Expression of moment of inertia along the x-axis.

$I_x=\frac{1}{4}m{R}^2$

Here, the radius of the disc is. $R$.

Expression of moment of inertia along y-axis and z-axis.

$I_y=\frac{1}{2}m{R}^2$

Expression of moment of inertia along z-axis.

$I_z=\frac{1}{4}m{R}^2$

Expression for conservation of linear momentum.

$F\left(\Delta t\right)j-m{\overline{v}}_{0}j=m{\overline{v}}_{x}i+m{\overline{v}}_{y}j+m{\overline{v}}_{z}k$ ...... (I)

Here, $\left(-ve\right)$ sign for $v_0$ shows the downward direction, velocity along the x-axis is $v_x$, velocity along y-axis is $v_y$, velocity along z-axis is $v_z$, and impulse at C is $F\left(\Delta t\right)$.

Compare x-component of Equation (I) on both side.

$\begin{array}{c}m{\overline{v}}_{x}i=0\\ {\overline{v}}_x=0\end{array}$

Compare y-component of Equation (I) on both side.

$\begin{array}{c}F\left(\Delta t\right)j-m{\overline{v}}_{0}j=m{\overline{v}}_{y}j\\ F\left(\Delta t\right)-m{\overline{v}}_0=m{\overline{v}}_y\\ F\left(\Delta t\right)=m\left({\overline{v}}_0+{\overline{v}}_y\right)\end{array}$ ...... (II)

Compare z-component of Equation (I) on both side.

$\begin{array}{c}m{\overline{v}}_{z}k=0\\ {\overline{v}}_z=0\end{array}$

Expression of relative position of C according to center of mass.

$R_c=\sqrt{\frac{1}{2}}R\left(i-k\right)$

As $e=0$, the velocity of circular disc is zero along y-axis after impact.

Expression of the velocity of circular disc C.

${\overline{v}}_c=\overline{v}+\omega \times R_c$ ...... (III)

Substitute $\left(v_{cx}i+v_{cy}j+v_{cz}k\right)$ for $v_{cx}$, $\left({\overline{v}}_{x}i+{\overline{v}}_{y}j+{\overline{v}}_{z}k\right)$ for $\overline{v}$, and $\left({\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k\right)$ for $\omega$ in the Equation (III).

$\begin{array}{l}\left(v_{cx}i+v_{cy}j+v_{cz}k\right)=\left({\overline{v}}_{x}i+{\overline{v}}_{y}j+{\overline{v}}_{z}k\right)+\left({\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k\right)\times R_c\\ \end{array}$ ...... (IV)

Substitute $0$ for ${\overline{v}}_x$, ${\overline{v}}_z$ and $\left(v_{cy}j\right)$, $\sqrt{\frac{1}{2}}R\left(i-k\right)$ for $R_c$ in Equation (IV)

$\begin{array}{c}{v}_{cx}i+v_{cz}k=\left({\overline{v}}_{y}j\right)+\left({\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k\right)\times \left(\sqrt{\frac{1}{2}}R\left(i-k\right)\right)\\ v_{cx}i+v_{cz}k=\left({\overline{v}}_{y}j\right)+\left(\sqrt{\frac{1}{2}}R\left(i-k\right){\omega }_{x}i+\sqrt{\frac{1}{2}}R\left(i-k\right){\omega }_{y}j+\sqrt{\frac{1}{2}}R\left(i-k\right){\omega }_{z}k\right)\\ v_{cx}i+v_{cz}k=\left({\overline{v}}_{y}j\right)+\left(\sqrt{\frac{1}{2}}R{\omega }_{x}j-\sqrt{\frac{1}{2}}R{\omega }_y\left(i+k\right)+\sqrt{\frac{1}{2}}R{\omega }_{z}j\right)\end{array}$ ...... (V)

Compare y-component of Equation (V) on both sides.

$\begin{array}{c}0=\left({\overline{v}}_{y}j\right)+\left(\sqrt{\frac{1}{2}}R{\omega }_{x}j+\sqrt{\frac{1}{2}}R{\omega }_{z}j\right)\\ {\overline{v}}_y=-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\end{array}$ ...... (VI)

Substitute $\left(-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)$ for ${\overline{v}}_y$ in Equation (II).

$F\left(\Delta t\right)=m\left({\overline{v}}_0-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)$ ...... (VII)

Expression of moment about center of mass.

${\overline{I}}_x{\omega }_{x}i+{\overline{I}}_y{\omega }_{y}j+{\overline{I}}_z{\omega }_{z}k=F\left(\Delta t\right)j\times R_c$ ...... (VIII)

Expression for kinetic energy of the circular plat before impact.

$T_0=\frac{1}{2}{I}_x{\omega }_x{}^2+\frac{1}{2}{I}_y{\omega }_y{}^2+\frac{1}{2}{I}_z{\omega }_z{}^2+\frac{1}{2}m{\overline{v}}^2$ ...... (IX)

Expression for kinetic energy of the circular plat after impact.

$T=\frac{1}{2}{I}_x{\omega }_x{}^2+\frac{1}{2}{I}_y{\omega }_y{}^2+\frac{1}{2}{I}_z{\omega }_z{}^2+\frac{1}{2}m{\overline{v}}^2$ ...... (X)

Expression of energy loss.

$U=T_0-T$ ...... (XI)

Calculation:

Substitute, $\frac{1}{4}m{R}^2$ for $I_x$, $\frac{1}{2}m{R}^2$ for $I_y$, $\frac{1}{4}m{R}^2$ and $I_z$, and $\sqrt{\frac{1}{2}}R\left(i-k\right)$ for $R_c$ in Equation (VIII).

$\left(\frac{1}{4}m{R}^2\right){\omega }_{x}i+\left(\frac{1}{2}m{R}^2\right){\omega }_{y}j+\left(\frac{1}{4}m{R}^2\right){\omega }_{z}k=F\left(\Delta t\right)\times \left(\sqrt{\frac{1}{2}}R\left(i+k\right)\right)$ ...... (XII)

Compare x-component of Equation (XII) on both side.

$\begin{array}{c}\left(\frac{1}{4}m{R}^2\right){\omega }_x=F\left(\Delta t\right)\times \left(\sqrt{\frac{1}{2}}R\right)\\ \left(\frac{\sqrt{2}}{4}mR\right){\omega }_x=F\left(\Delta t\right)\\ \left(\frac{1}{2\sqrt{2}}mR\right){\omega }_x=F\left(\Delta t\right)\\ F\left(\Delta t\right)=\left(\frac{1}{2\sqrt{2}}mR\right){\omega }_x\end{array}$

Compare y-component of Equation (XII) on both side.

$\begin{array}{c}\left(\frac{1}{2}m{R}^2\right){\omega }_y=0\\ {\omega }_y=0\end{array}$

Compare z-component of Equation (XII) on both side.

$\begin{array}{c}\left(\frac{1}{4}m{R}^2\right){\omega }_{z}k=F\left(\Delta t\right)\times \left(\sqrt{\frac{1}{2}}Rk\right)\\ \left(\frac{1}{4}mR\right){\omega }_z=F\left(\Delta t\right)\times \left(\sqrt{\frac{1}{2}}\right)\\ F\left(\Delta t\right)=\left(\frac{1}{2\sqrt{2}}mR{\omega }_z\right)\end{array}$

Substitute $\left(\frac{1}{2\sqrt{2}}mR\right){\omega }_x$ for $F\left(\Delta t\right)$ in Equation (VII).

$\begin{array}{c}\left(\frac{1}{2\sqrt{2}}mR\right){\omega }_x=m\left({\overline{v}}_0-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)\\ {\omega }_x=\frac{\left({\overline{v}}_0-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)}{\left(\frac{1}{2\sqrt{2}}R\right)}\\ =\frac{2\left(\sqrt{2}{\overline{v}}_0-R\left({\omega }_x+{\omega }_z\right)\right)}{R}\\ =\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2\left({\omega }_x+{\omega }_z\right)\right)\end{array}$

$\begin{array}{c}{\omega }_x=\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2\left({\omega }_x+{\omega }_z\right)\right)\\ {\omega }_x=\frac{\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2{\omega }_z\right)}{3}\end{array}$ ...... (XIII)

Substitute $\left(\frac{1}{2\sqrt{2}}mR{\omega }_z\right)$ for $F\left(\Delta t\right)$ in Equation (VII).

$\begin{array}{c}\left(\frac{1}{2\sqrt{2}}mR{\omega }_z\right)=m\left({\overline{v}}_0-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)\\ {\omega }_z=\frac{\left({\overline{v}}_0-\sqrt{\frac{1}{2}}R\left({\omega }_x+{\omega }_z\right)\right)}{\left(\frac{1}{2\sqrt{2}}R\right)}\\ =\frac{2\left(\sqrt{2}{\overline{v}}_0-R\left({\omega }_x+{\omega }_z\right)\right)}{R}\\ =\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2\left({\omega }_x+{\omega }_z\right)\right)\end{array}$

$\begin{array}{c}{\omega }_z=\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2\left({\omega }_x+{\omega }_z\right)\right)\\ {\omega }_z=\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2{\omega }_x-2{\omega }_z\right)\\ {\omega }_x=\frac{\frac{2\sqrt{2}{\overline{v}}_0}{R}-3{\omega }_z}{2}\end{array}$ ...... (XIV)

Equate Equation (XIII) and Equation (XIV).

$\begin{array}{c}\frac{\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-2{\omega }_z\right)}{3}=\frac{\left(\frac{2\sqrt{2}{\overline{v}}_0}{R}-3{\omega }_z\right)}{2}\\ \left(\frac{4\sqrt{2}{\overline{v}}_0}{R}-4{\omega }_z\right)=\left(\frac{6\sqrt{2}{\overline{v}}_0}{R}-9{\omega }_z\right)\\ \frac{2\sqrt{2}{\overline{v}}_0}{R}-5{\omega }_z=0\\ {\omega }_z=\frac{2\sqrt{2}{\overline{v}}_0}{5R}\end{array}$

Substitute $\frac{2\sqrt{2}{\overline{v}}_0}{5R}$ for ${\omega }_z$ in Equation (XIV).

$\begin{array}{l}{\omega }_x=\frac{\frac{2\sqrt{2}{\overline{v}}_0}{R}-\frac{6\sqrt{2}{\overline{v}}_0}{5R}}{2}\\ {\omega }_x=\frac{\sqrt{2}{\overline{v}}_0}{R}-\frac{3\sqrt{2}{\overline{v}}_0}{5R}\\ {\omega }_x=\frac{2\sqrt{2}{\overline{v}}_0}{5R}\end{array}$

Substitute $\frac{2\sqrt{2}{\overline{v}}_0}{5R}$ for ${\omega }_x$ and $\frac{2\sqrt{2}{\overline{v}}_0}{5R}$ for ${\omega }_z$ in Equation (VI)

$\begin{array}{c}{\overline{v}}_y=-\sqrt{\frac{1}{2}}R\left(\frac{2\sqrt{2}{\overline{v}}_0}{5R}+\frac{2\sqrt{2}{\overline{v}}_0}{5R}\right)\\ =-\sqrt{\frac{1}{2}}R\left(\frac{4\sqrt{2}{\overline{v}}_0}{5R}\right)\\ =-\frac{4}{5}{\overline{v}}_0\end{array}$

Expression for velocity along x-axis, y-axis and z-axis.

$\overline{v}=\left({\overline{v}}_{x}i+{\overline{v}}_{y}j+{\overline{v}}_{z}k\right)$ ...... (XV)

Substitute 0 for ${\overline{v}}_x$, 0 for ${\overline{v}}_y$, and $-\frac{4}{5}{\overline{v}}_0$ for ${\overline{v}}_z$ in Equation (XV).

$\overline{v}=-\frac{4}{5}{\overline{v}}_{0}j$

Substitute $0$ for ${\omega }_x$, ${\omega }_z$ ,and ${\omega }_y$, $\frac{1}{4}m{R}^2$ for $I_x$, $\frac{1}{2}m{R}^2$ for $I_y$, $\overline{v}$ for ${\overline{v}}_0$ and $\frac{1}{4}m{R}^2$ for $I_z$ in Equation (IX)

$\begin{array}{c}{T}_0=\frac{1}{2}\left(\frac{1}{4}m{R}^2\right){\left(0\right)}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(0\right)}^2+\frac{1}{2}\left(\frac{1}{4}m{R}^2\right){\left(0\right)}^2+\frac{1}{2}m{\left({\overline{v}}_0\right)}^2\\ =\frac{m{\overline{v}}_0{}^2}{2}\end{array}$

Substitute $\frac{2\sqrt{2}{\overline{v}}_0}{5R}$ for ${\omega }_x$, $\frac{2\sqrt{2}{\overline{v}}_0}{5R}$ for ${\omega }_z$, $0$ for ${\omega }_y$, $\frac{1}{4}m{R}^2$ for $I_x$, $\frac{1}{2}m{R}^2$ for $I_y$, $-\frac{4}{5}{\overline{v}}_{0}j$ for, and $\frac{1}{4}m{R}^2$ for $I_z$ in Equation (X)

$\begin{array}{c}T=\frac{1}{2}\left(\frac{1}{4}m{R}^2\right){\left(\frac{2\sqrt{2}{\overline{v}}_0}{5R}\right)}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(0\right)}^2+\frac{1}{2}\left(\frac{1}{4}m{R}^2\right){\left(\frac{2\sqrt{2}{\overline{v}}_0}{5R}\right)}^2+\frac{1}{2}m{\left(-\frac{4}{5}{\overline{v}}_{0}j\right)}^2\\ T=\left(\frac{1}{4}m\right)\left(\frac{4{\overline{v}}_0{}^2}{25}\right)+\left(\frac{1}{4}m\right)\left(\frac{4{\overline{v}}_0{}^2}{25}\right)+m{\left(-\frac{4}{5}{\overline{v}}_{0}j\right)}^2.\\ =\frac{m{\overline{v}}_0{}^2}{25}+\frac{m{\overline{v}}_0{}^2}{25}+\frac{8m{\overline{v}}_0{}^2}{25}\\ =\frac{10m{\overline{v}}_0{}^2}{25}\end{array}$

$T=\frac{2m{\overline{v}}_0{}^2}{5}$

Substitute $\frac{m{\overline{v}}_0{}^2}{2}$ for $T_0$ and $\frac{2m{\overline{v}}_0{}^2}{5}$ for $T$ in Equation (XI).

$\begin{array}{c}U=\left(\frac{m{\overline{v}}_0{}^2}{2}\right)-\left(\frac{2m{\overline{v}}_0{}^2}{5}\right)\\ =\frac{m{\overline{v}}_0{}^2}{10}\end{array}$

Conclusion:

Thus, the energy loss of circular disc after impact is $\frac{m{\overline{v}}_0{}^2}{10}$.