Chapter 17.1, Problem 17.12P

Solve Prob. 17.11, assuming that the 6 N $\cdot$ m couple is applied to gear B.

To determine

(a)

Number of revolutions of gear C.

Answer to Problem 17.12P

Number of revolutions of gear C is ${\theta }_C=58.07rev$

Explanation of Solution

Given information:

Mass of the gear A (m_A) = 10kg.

Radius of gyration of the gear A (k_A) = 190mm.

Mass of the gear B (m_B) = 10kg.

Radius of gyration of the gear B (k_B) = 190mm.

Mass of the gear C (m_C) = 2.5kg.

Radius of gyration of the gear C (k_C) = 80mm.

Initial angular velocity of gear C (N_c)₁ = 450rpm

Final angular velocity of gear C (N_C)₁ = 1800rpm

A couple at gear C (M) = 6N-m.

Radius of gear A (r_A) = 250mm

Radius of gear B (r_B) = 250mm

Radius of gear C (r_C) = 100mm

Calculation:

Moment of inertia of gear A

$\begin{array}{l}{I}_A=m_A{k}_A{}^2\\ I_A=10\times {0.190}^2\\ I_A=0.361kg-m^2\end{array}$

Moment of inertia of gear B

$\begin{array}{l}{I}_B=m_B{k}_B{}^2\\ I_B=10\times {0.190}^2\\ I_B=0.361kg-m^2\end{array}$

Moment of inertia of gear C

$\begin{array}{l}{I}_C=m_C{k}_C{}^2\\ I_C=2.5\times {0.080}^2\\ I_C=0.016kg-m^2\end{array}$

For Initial condition; angular velocity is 450 rpm

Angular velocity of the gear C,

$\begin{array}{l}{\omega }_C=\frac{2\pi N_C}{60}\\ {\omega }_C=\frac{2\pi \times 450}{60}\\ {\omega }_C=47.1radian/s\end{array}$

Gear C mess with Gear A. So angular velocity ration is given as

$\begin{array}{l}\frac{{\omega }_C}{{\omega }_A}=\frac{r_A}{r_C}\\ \frac{47.1}{{\omega }_A}=\frac{250}{100}\\ {\omega }_A=\frac{100}{250}\times 47.1\\ {\omega }_A=18.84radian/s\end{array}$

Gear C mess with gear B

$\begin{array}{l}\frac{{\omega }_B}{{\omega }_C}=\frac{r_C}{r_B}\\ \frac{{\omega }_B}{47.1}=\frac{100}{250}\\ {\omega }_B=\frac{100}{250}\times 47.1\\ {\omega }_B=18.84radian/s\end{array}$

Initial kinetic energy

$\begin{array}{l}{E}_1=\frac{1}{2}\left(I_A{\omega }_A{}^2+I_B{\omega }_B{}^2+I_C{\omega }_C{}^2\right)\\ E_1=\frac{1}{2}\times \left(0.361\times {18.84}^2+0.361\times {18.84}^2+0.016\times {47.1}^2\right)\\ E_1=\frac{1}{2}\times \left(128.135+128.135+35.495\right)\\ E_1=145.882J\end{array}$

For final condition; Angular velocity of the gear C is 1800rpm

$\begin{array}{l}{\omega }_C=\frac{2\pi N_C}{60}\\ {\omega }_C=\frac{2\pi \times 1800}{60}\\ {\omega }_C=188.4radian/s\end{array}$

Gear C mess with Gear A. So angular velocity ration is given as

$\begin{array}{l}\frac{{\omega }_C}{{\omega }_A}=\frac{r_A}{r_C}\\ \frac{188.4}{{\omega }_A}=\frac{250}{100}\\ {\omega }_A=\frac{100}{250}\times 188.4\\ {\omega }_A=75.36radian/s\end{array}$

Gear C mess with gear B

$\begin{array}{l}\frac{{\omega }_B}{{\omega }_C}=\frac{r_C}{r_B}\\ \frac{{\omega }_B}{188.4}=\frac{100}{250}\\ {\omega }_B=\frac{100}{250}\times 188.4\\ {\omega }_B=75.36radian/s\end{array}$

Final kinetic energy

$\begin{array}{l}{E}_2=\frac{1}{2}\left(I_A{\omega }_A{}^2+I_B{\omega }_B{}^2+I_C{\omega }_C{}^2\right)\\ E_2=\frac{1}{2}\times \left(0.361\times {75.36}^2+0.361\times {75.36}^2+0.016\times {188.4}^2\right)\\ E_2=\frac{1}{2}\times \left(2050.166+2050.166+567.913\right)\\ E_2=2334.122J\end{array}$

Work done by the gear C

$\begin{array}{l}Work=couple\times angle\\ W=M\times {\theta }_C\\ W=6\times {\theta }_C\end{array}$

Substitute the value of E₁, E₂ and W in work energy equation

$\begin{array}{l}{E}_1+W=E_2\\ 145.882+6{\theta }_C=2334.122\\ 6{\theta }_C=2188.240\\ {\theta }_C=\frac{2188.240}{6}\\ {\theta }_C=364.70radian\\ {\theta }_C=364.70\times \frac{1}{2\pi }\\ {\theta }_C=58.07rev\end{array}$

To determine

(b)

Tangential force on gear A.

Answer to Problem 17.12P

Tangential force on gear A is $F=0.0447N$

Explanation of Solution

Given information:

Mass of the gear A (m_A) = 10kg.

Radius of gyration of the gear A (k_A) = 190mm.

Mass of the gear B (m_B) = 10kg.

Radius of gyration of the gear B (k_B) = 190mm.

Mass of the gear C (m_C) = 2.5kg.

Radius of gyration of the gear C (k_C) = 80mm.

Initial angular velocity of gear C (N_c)₁ = 450rpm

Final angular velocity of gear C (N_C)₁ = 1800rpm

A couple at gear C (M) = 6N-m.

Radius of gear A (r_A) = 250mm

Radius of gear B (r_B) = 250mm

Radius of gear C (r_C) = 100mm

Calculation:

Angle of rotation for gear A

$\begin{array}{l}\frac{{\theta }_A}{{\theta }_C}=\frac{{\omega }_C}{{\omega }_A}\\ \frac{{\theta }_A}{364.70}=\frac{188.4}{75.36}\\ {\theta }_A=911.75radian\end{array}$

Substitute the value of E₁, E₂ and W in work energy equation for gear A

$\begin{array}{l}{\left(E_1\right)}_A+{\left(W\right)}_A={\left(E_2\right)}_A\\ \frac{1}{2}{I}_A{\left({\omega }_1\right)}_A+M{\theta }_A=\frac{1}{2}{I}_A{\left({\omega }_2\right)}_A\\ \frac{1}{2}\times 0.361\times 18.84+\left(F\times r_A\right)\times 911.75=\frac{1}{2}\times 0.361\times 75.36\\ 3.4+F\times 0.25\times 911.75=13.6\\ F\times 227.9=10.2\\ F=\frac{10.2}{227.9}\\ F=0.0447N\end{array}$