Chapter 17.2, Problem 17.89P

To determine

Calculate the angular velocity of frame and moment of inertia of pulley and frame about rod CD.Coller A and B slide on frame collars are attached by a cord to slide over a frame.

Answer to Problem 17.89P

Angular velocity of the frame is $18.83rad/sec$ and moment of inertia of pully and frame is $0.0508kg.m^2$

Explanation of Solution

Given information:

Mass of coller A = $m_A=1.8kg.$

Mass of coller B = $m_B=0.7kg.$ The velocity of coller A = $\left(v_A\right)=2.1m/s$

$\begin{array}{l}{\left(r_A\right)}_1=0.1m\hfill \\ {\left(r_A\right)}_2=0.12m\hfill \end{array}$

The velocity of coller A at position $2=\left(v_A\right)2=2.5m/s$

Concept used:

Conservation of angular momentum.

Conservation of energy.

calculation:

component of velocity for coller A,

${(v_A)}^2={(v_A)}_r{}^2+{(v_A)}_{\theta }{}^2$ ---------------------(1)

${(v_A)}_{\theta }=r_A\omega$ ---------------------rod constraint

$\begin{array}{l}\Delta r_A=\Delta \gamma \beta ,\\ {(}_{v_A}=v_B\end{array}$ ---------------------cable constraint

For position 1,

$\begin{array}{l}(r_A)=0.1\\ $ {(v_A)}_1=2.1m/s$$ {[{(v_A)}_r]}_1=0$\end{array}$

From equation (1),

$\begin{array}{l}$ {(v_A)}_{\theta }=2.1$2.1=0.1\omega ,\\ {\omega }_1=21rad/\sec \\ v_B=0\end{array}$

Potential energy, $v_1=0$

Kinetic energy, $E_1=\frac{I{\omega }_1{}^2}{2}+\frac{m_A{v}_A{}^2}{2}+\frac{m_B{v}_B{}^2}{2}$

Angular momentum,

$\begin{array}{l}$ {(H_0)}_1=I{\omega }_1+m_A{[{(v_A)}_r]}_1{(r_A)}_1$=21I+0.378\end{array}$

For position 2,

$\begin{array}{l}$ {(r_A)}_2=0.12$$ {(v_A)}_2=2.5m/s$\omega ={\omega }_2\\ $ {[{(v_A)}_{\theta }]}_2=0.12\omega 2$$ {[{(v_A)}_r]}_2{}^2={[{(v_A)}_2]}^2-{[{(v_A)}_{\theta }]}_2{}^2$={2.5}^2-0.0144{\omega }_2{}^2\\ =6.25-0.0144{\omega }_2{}^2\\ v_B{}^2=6.25-0.0144{\omega }_2{}^2\\ $ \Delta r_A={(r_A)}_2-{(r_A)}_1$=0.02m\\ \Delta Y_B\end{array}$

Potential energy,

$\begin{array}{l}{v}_2=m_{B}g(\Delta Y_B)\\ v_2=0.13734J\end{array}$

Kinetic energy,

$\begin{array}{l}{E}_2=\frac{I{\omega }_2{}^2}{2}+\frac{m_A{v}_A{}^2}{2}+\frac{m_B{v}_B{}^2}{2}\\ E_2=(0.5I-0.00504){\omega }_2{}^2+7.8125\end{array}$

Angular momentum,

$\begin{array}{l}$ {(H_0)}_2=I{\omega }_2+m_A{[{(v_A)}_{\theta }]}_2{(r_A)}_2$=(I+0.0259){\omega }_2\end{array}$

By conservation of angular momentum,

$\begin{array}{l}$ {(H_0)}_1={(H_0)}_2$21I+0.378=(I+0.0259){\omega }_2\\ {\omega }_2=\frac{21I+0.378}{I+0.0259}=\frac{A}{B}\end{array}$

By conservation of energy,

$\begin{array}{l}{E}_1+v_1=E_2+v_2\\ =220.5I+3.969+0\\ =(0.5I-0.00504){\omega }_2{}^2+7.8125+0.1373\\ 220.5I=(0.5I-0.00504)(\frac{A^2}{B^2})+3.9808\\ 220.5I-(0.5I-0.00504)(\frac{A^2}{B^2})-3.9808=0\\ 220.5I{B}^2-0.5I{A}^2+0.00504A^2-3.9808B^2=0\\ $ 220.5I{(I+0.0259)}^2-0.5I{(21I+0.378)}^{2}0.00504{(21I+0.378)}^2-3.9808{(I+0.0259)}^2=0$\end{array}$

Further solving we get

$\begin{array}{l}o{I}^3+1.7345I^2-0.04965I-0.001954=0\\ I=0.0507and\\ I=-0.0221\end{array}$

Considering only the positive value.

Hence,

Moment of inertia of pulley and frame is 0.0507 kg.m²

To find the angular velocity of the frame, we have

$\begin{array}{l}{\omega }_2=\frac{21I+0.378}{I+0.0259}\\ =\frac{21\times 0.0507+0.378}{0.0507+0.0259}\\ {\omega }_2=18.83rd/\sec \end{array}$

Conclusion:

By applying conservation of angular momentum and conservation of energy for initial and find the position of coller relative to the frame we get,

Angular velocity of the frame is $18.83rad/sec$ and moment of inertia of the system is $0.0507$ kgm².