Chapter 17.1, Problem 17.11P

Each of the gears A and B has a mass of 10 kg and a radius of gyration of 190 mm, while gear C has a mass of 2.5 kg and a radius of gyration of 80 mm. If a couple M of constant magnitude 6 N $\cdot$ m is applied to gear C, determine (a) the number of revolutions of gear C required for its angular velocity to increase from 450 rpm to 1800 rpm, (b) the corresponding tangential force acting on gear A.

  

To determine

(a)

Number of revolutions of gear C.

Answer to Problem 17.11P

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 of the gear C is 450 rpm,

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

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 17.11P

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

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