Chapter 17.1, Problem 17.17P

To determine

(a)

The distance b for which the angular velocity of the rod is maximum when it passes through a vertical position.

Answer to Problem 17.17P

The distance b: $b=\frac{l}{\sqrt{12}}$

Explanation of Solution

Given information:

Length of the slender rod is l. The distance between the pivot and the center of rod is b.

The rod is released from rest in a horizontal position and swings freely.

Calculation:

$\begin{array}{l}\underset{\_}{\text{At Position }\text{1}\text{.}}\\ \overline{v}=0,\\ \omega =0\\ \therefore K\text{i}\text{n}eticenergy:T_1=0\\ Elevation:\\ h=0\\ \therefore Potentialenergy:V_1=mgh=0\\ \underset{\_}{\text{At Position }\text{2}\text{.}}\\ {\overline{v}}_2=b{\omega }_2\\ I=\frac{1}{12}m{l}^2\\ \therefore K\text{i}\text{n}eticenergy:T_2=\frac{1}{12}m{\overline{v}}_2^2+\frac{1}{12}\overline{I}{\omega }_2^2\\ T_2=\frac{1}{12}m\left(b^2+\frac{1}{12}{l}^2\right){\omega }_2^2\\ Elevation:\\ h=-b\\ \therefore Potentialenergy:V_2=-mgb\\ \text{Applying the principle of conservation of energy:}\\ T_1+V_1=T_2+V_2:\\ 0+0=\frac{1}{2}m\left(b^2+\frac{1}{12}{l}^2\right){\omega }_2^2-mgb\\ {\omega }_2^2=\frac{2gb}{b^2+\frac{1}{12}{l}^2}\mathrm{................}\left(1\right)\end{array}$

$\begin{array}{l}\text{For maximum value of }{\omega }_2;\\ \frac{d}{db}\left({\omega }_2\right)=0\\ \frac{d}{db}\left(\frac{b}{b^2+\frac{1}{12}{l}^2}\right)=\frac{\left(b^2+\frac{1}{12}{l}^2\right)-b\left(2b\right)}{\left(b^2+\frac{1}{12}{l}^2\right)}=0\\ b^2=\frac{1}{12}{l}^2\\ \Rightarrow b=\frac{l}{\sqrt{12}}\end{array}$

Conclusion:

For $b=\frac{l}{\sqrt{12}}$, the angular velocity of the rod is maximum when it passes through a vertical position.

To determine

(b)

The corresponding value of the angular velocity and the reaction at C.

Answer to Problem 17.17P

The corresponding value of the angular velocity: ${\omega }_2=1.861\sqrt{\frac{g}{l}}$

The reaction at C: $C=2mg$

Explanation of Solution

Given information:

Length of the slender rod is l. The distance between the pivot and the center of rod is b.

The rod is released from rest in a horizontal position and swings freely.

From the result of part (a), $b=\frac{l}{\sqrt{12}}$.

Calculations:

$\begin{array}{l}\text{Using eq}\text{.}\left(\text{1}\right)\text{, from part (a), The angular velocity of the rod when it passes through }\\ \text{a vertical position is:}\\ {\omega }_2={\left(\frac{2gb}{b^2+\frac{1}{12}{l}^2}\right)}^{\frac{1}{2}}\\ substitut\text{i}\text{n}g,b=\frac{l}{\sqrt{12}}\\ {\omega }_2={\left(\frac{2g\frac{l}{\sqrt{12}}}{\frac{l^2}{12}+\frac{l^2}{12}}\right)}^{\frac{1}{2}}\\ {\omega }_2^2=\sqrt{12}\frac{g}{l}\\ {\omega }_2={12}^{1/4}\sqrt{\frac{g}{l}}\\ \Rightarrow {\omega }_2=1.861\sqrt{\frac{g}{l}}\end{array}$

$\begin{array}{l}Fromthefreebodyandk\text{i}\text{n}eticdiagramoftherodwhenitpassesthroughthe\\ verticalposition\left(asshownabove\right):\\ {\text{a}}_n=b{\omega }_2^2\\ {\text{a}}_n=\frac{l}{\sqrt{12}}\sqrt{12}\frac{g}{l}\\ {\text{a}}_n=g\\ \text{Applying force balance in y-direction:}\\ +\uparrow \sum F_y=m{\text{a}}_n:\\ C_y-mg=mg\\ C_y=2mg\\ \sum M_C=0:\\ mb{a}_t+\overline{I}\alpha =0\\ with,\left(a_t=b\alpha \right)\\ \left(m{b}^2+\overline{I}\right)\alpha =0\\ \alpha =0,\\ hence,a_t=0\\ \text{Applying force balance in x-direction:}\\ \sum F_x=m{a}_t:\\ C_x=-m{a}_t=0\\ \text{Thus, the reaction at C:}\\ \Rightarrow C=2mg\end{array}$

Conclusion:

The corresponding value of the angular velocity is ${\omega }_2=1.861\sqrt{\frac{g}{l}}$ and the reaction at C is $C=2mg$.