Chapter 17.1, Problem 17.35P

The 1.5-kg uniform slender bar AB is connected to the 3-kggear B that meshes with the stationary outer gear C. The centroidal radius of gyration of gear B is 30 mm. Knowing that the system is released from rest in the position shown, determine (a) the angular velocity of the bar as it passes through the vertical position, (b) the corresponding angular velocity of gear B.

  

To determine

i.

Calculate the angular velocity of the bar when the bar is a vertical position.

Answer to Problem 17.35P

Angular velocity of the bar AB is $11.57rad/\sec$ a vertical position.

Explanation of Solution

Given:

Bar AB is connected to gear B which meshes with outer gear C. System is released from the rest.

Mass of bar AB, $m_{AB}=1.5kg$

Mass of gear B, $m_B=3kg$

The radius of gyration of gear B, $k_B=30mm=0.03m$

The radius of gear B, $r_B=50mm=0.05m$

Length of Bar AB, $L_{AB}=120mm=0.12m$

Concept used:

Energy conservation principle

Calculation:

Initial Position: the system is at rest

Kinetic energy, $T_1=\frac{m{v}_1{}^2}{2}=0(v_1=0)$

Potential energy, $v_1=mg{h}_1=0(h_1=0)$

Final position: the bar is a vertical position

Moment of inertia of bar AB,

$I_{AB}=\frac{m_{AB}{L}_{AB}{}^2}{12}=0.0018kg.m^2$

Moment of inertia of gear B,

$I_B=m_B{k}_B{}^2=0.0027kg.m^2$

$v_B=L_{AB}{\omega }_{AB}$

But, $v_B=r_B{\omega }_B$

${\omega }_B=\frac{L_{AB}}{r_B}{\omega }_{AB}$

Kinetic energy,

$\begin{array}{l}T2=\frac{m_{AB}{v}_{AB}{}^2}{2}+\frac{I_{AB}{\omega }_{AB}{}^2}{2}+\frac{m_B{v}_B{}^2}{2}+\frac{I_B{\omega }_B{}^2}{2}\\ =\frac{m_{AB}{L}_{AB}{}^2{\omega }_{AB}{}^2}{8}+\frac{I_{AB}{\omega }_{AB}{}^2}{2}+\frac{m_B{L}_{AB}{}^2{\omega }_{AB}{}^2}{2}+\frac{I_B{L}_{AB}{}^2{\omega }_{AB}{}^2}{2r_B{}^2}\\ =0.0027{\omega }_{AB}{}^2+9\times {10}^{-4}{\omega }_{AB}{}^2+0.0216{\omega }_{AB}{}^2+7.776\times {10}^{-3}{\omega }_{AB}{}^2\\ =0.03297{\omega }_{AB}{}^2\end{array}$

Potential energy,

$\begin{array}{l}{v}_2=-w_{AB}\frac{L_{AB}}{2}-w_B{L}_{AB}\\ =(-m_{AB}\times g)0.06-(m_B\times g)0.12\\ =-0.8829-3.5316\\ =-4.4145J\end{array}$

By energy conservation principle;

$\begin{array}{l}{T}_1+v_1=T_2+v_2\\ 0=0.03297{\omega }_{AB}{}^2-4.4145\\ {\omega }_{AB}=11.57rad/\sec \end{array}$

Conclusion:

Thus, the angular velocity of rod AB can be calculated by the energy conservation principle by considering the initial and final condition of the system.

To determine

ii.

Calculate the angular velocity of gear B.

Answer to Problem 17.35P

Angular velocity of gear B is $27.768rad/\sec$.

Explanation of Solution

Given:

Mass of bar AB, $m_{AB}=1.5kg$

Mass of gear B, $m_B=3kg$

The radius of gyration of gear B, $k_B=30mm=0.03m$

The radius of gear B, $r_B=50mm=0.05m$

Length of Bar AB, $L_{AB}=120mm=0.12m$

Concept used:

Energy conservation principle

Calculation:

Considering the value calculated in sub-part (i);

${\omega }_{AB}=11.57rad/\sec$

We know that,

$\begin{array}{l}{\omega }_B=\frac{L_{AB}}{r_B}{\omega }_{AB}\\ =\frac{0.12}{0.05}11.57\\ {\omega }_B=27.768rad/\sec \end{array}$

Conclusion:

Thus, the corresponding angular velocity of gear B can be calculated by simple calculations.