Chapter 17.3, Problem 17.125P

To determine

(a)

Velocity of the block and angular velocity of the disk immediately after perfectly plastic impact.

Answer to Problem 17.125P

The velocity of the block and angular velocity of the disk immediately after perfectly plastic impactare $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}M+m\right)}$ and $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}MR+mR\right)}$ respectively.

Explanation of Solution

Given:

Mass of block is $m$, mass of the disk is $M$ and Falling distance is $h$.

Concept used:

Conservation of momentum for perfectly plastic condition is given as follows:

$I{\omega }_1+m{u}_{1}R=I{\omega }_2+m{u}_{2}R$ …… (1)

Here, angular velocity of disk is $\omega$ and velocity of the blockis $u$. 1 and 2 subscripts are for initial and final conditions.

Calculation:

Velocity of the block after the cord becomes taut is calculated by conservation of energy as follows:

$\begin{array}{c}mgh=\frac{1}{2}m{u}^2\\ u=\sqrt{2gh}\end{array}$

Substitute $0$ for ${\omega }_1$, $\sqrt{2gh}$ for $u_1$, $\frac{1}{2}M{R}^2$ for $I$ and ${\omega }_{2}R$ for $u_2$ in equation (1).

$\begin{array}{c}I\times 0+m\times \sqrt{2gh}\times R=\frac{1}{2}M{R}^2\times {\omega }_2+m\times {\omega }_2{R}^2\\ m\times \sqrt{2gh}={\omega }_2\left(\frac{1}{2}MR+mR\right)\\ {\omega }_2=\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}MR+mR\right)}\end{array}$

Velocity of the block is calculated as follows:

$\begin{array}{l}{u}_2={\omega }_{2}R\\ u_2=\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}MR+mR\right)}\times R\\ u_2=\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}M+m\right)}\end{array}$

Thus, the velocity of the block and angular velocity of the disk immediately after perfectly plastic impactare $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}M+m\right)}$ and $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}MR+mR\right)}$ respectively.

Conclusion:

The velocity of the block and angular velocity of the disk immediately after perfectly plastic impactare $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}M+m\right)}$ and $\frac{m\times \sqrt{2gh}}{\left(\frac{1}{2}MR+mR\right)}$ respectively.

To determine

(b)

Velocity of the block and angular velocity of the disk immediately after perfectly elastic impact.

Answer to Problem 17.125P

The velocity of the block and angular velocity of the disk immediately after perfectly plastic impactare $\frac{4m\sqrt{2gh}}{R\left(2m+M\right)}$ and $\sqrt{2gh}\left(1-\frac{2M}{\left(2m+M\right)}\right)$ respectively.

Explanation of Solution

Given:

Mass of block is $m$, mass of the disk is $M$ and Falling distance is $h$.

Calculation:

Substitute $0$ for ${\omega }_1$, $\sqrt{2gh}$ for $u_1$ and $\frac{1}{2}M{R}^2$ for $I$ in equation (1).

$\begin{array}{c}I\times 0+m\times \sqrt{2gh}\times R=\frac{1}{2}M{R}^2\times {\omega }_2+m\times u_{2}R\\ u_2=\frac{mR\sqrt{2gh}-\frac{1}{2}M{R}^2\times {\omega }_2}{mR}\\ u_2=\sqrt{2gh}-\frac{1}{2}\times \frac{MR}{m}\times {\omega }_2\end{array}$

Apply the conservation of energy for elastic impact as follows:

$mgh+\frac{1}{2}I{\omega }_1{}^2=\frac{1}{2}I{\omega }_2{}^2+\frac{1}{2}m{u}_2{}^2$

Substitute $0$ for ${\omega }_1$, $\sqrt{2gh}-\frac{1}{2}\times \frac{MR}{m}\times {\omega }_2$ for $u_2$ and $\frac{1}{2}M{R}^2$ for $I$ in above equation.

$\begin{array}{l}mgh+\frac{1}{2}I{\left(0\right)}^2=\frac{1}{2}\times \frac{1}{2}M{R}^2{\omega }_2{}^2+\frac{1}{2}m{\left(\sqrt{2gh}-\frac{1}{2}\times \frac{MR}{m}\times {\omega }_2\right)}^2\\ mgh=\frac{1}{4}M{R}^2{\omega }_2{}^2+\frac{1}{2}m\left(2gh-2\times \sqrt{2gh}\times \frac{1}{2}\times \frac{MR}{m}\times {\omega }_2+\frac{1}{4}\times \frac{M^2{R}^2}{m^2}\times {\omega }_2{}^2\right)\\ mgh=\frac{1}{4}M{R}^2{\omega }_2{}^2+\left(mgh-\sqrt{2gh}\times \frac{1}{2}\times MR\times {\omega }_2+\frac{1}{8}\times \frac{M^2{R}^2}{m}\times {\omega }_2{}^2\right)\\ \left(\frac{1}{4}M{R}^2+\frac{M^2{R}^2}{8m}\right){\omega }_2{}^2-\sqrt{2gh}\times \frac{MR}{2}\times {\omega }_2+mgh-mgh=0\\ {\omega }_2\left(\left(\frac{1}{4}M{R}^2+\frac{M^2{R}^2}{8m}\right){\omega }_2-\sqrt{2gh}\times \frac{MR}{2}\right)=0\end{array}$

On further simplification,

$\begin{array}{l}\left(\left(\frac{1}{4}M{R}^2+\frac{M^2{R}^2}{8m}\right){\omega }_2-\sqrt{2gh}\times \frac{MR}{2}\right)=0\\ {\omega }_2=\frac{\sqrt{2gh}\times \frac{MR}{2}}{\left(\frac{1}{4}M{R}^2+\frac{M^2{R}^2}{8m}\right)}\\ {\omega }_2=\frac{4m\sqrt{2gh}}{R\left(2m+M\right)}\end{array}$

Velocity of the block is calculated as follows:

$\begin{array}{l}{u}_2=\sqrt{2gh}-\frac{1}{2}\times \frac{MR}{m}\times \frac{4m\sqrt{2gh}}{R\left(2m+M\right)}\\ u_2=\sqrt{2gh}\left(1-\frac{2M}{\left(2m+M\right)}\right)\end{array}$

Thus, the velocity of the block and angular velocity of the disk immediately after perfectly elastic impactare $\frac{4m\sqrt{2gh}}{R\left(2m+M\right)}$ and $\sqrt{2gh}\left(1-\frac{2M}{\left(2m+M\right)}\right)$ respectively.

Conclusion:

The velocity of the block and angular velocity of the disk immediately after perfectly elastic impactare $\frac{4m\sqrt{2gh}}{R\left(2m+M\right)}$ and $\sqrt{2gh}\left(1-\frac{2M}{\left(2m+M\right)}\right)$ respectively.