# Gear and Gear Train

 Question 1
Two meshing spur gears 1 and 2 with diametral pitch of 8 teeth per mm and an angular velocity ratio $|\omega _2|/|\omega _1|=1/4$, have their centers 30 mm apart. The number of teeth on the driver (gear 1) is _______ . (Answer in integer) A 52 B 48 C 96 D 112
GATE ME 2023   Theory of Machine
Question 1 Explanation: Given
$R+r=30 \quad \text {...(i) }$
$\frac{T_{1}}{T_{2}}=\frac{1}{4} ; \quad$ where $T=$ no. of teeth
or $\mathrm{T}_{2}=4 \mathrm{T}_{1}$
Diametrical pitch $=\frac{8 \text { teeth }}{\mathrm{mm}}=\frac{\mathrm{T}}{\mathrm{D}}=\mathrm{P}_{\mathrm{d}}$
and $m=\frac{D}{T}=\frac{1}{P_{d}}$
As, \begin{aligned} r & =\frac{m T}{2} \\ R & =\frac{m T_{2}}{2}, r=\frac{m T_{1}}{2} \end{aligned}

by equation (i) we have
$\frac{m T_{2}}{2}+\frac{m T_{1}}{2}=30$
or, $\frac{m}{2}\left(T_{1}+T_{2}\right)=30$
or, $\frac{1}{8 \times 2}\left(T_{1}+T_{2}\right)=30$
As $\mathrm{T}_{2}=4 \mathrm{T}_{1}$
So, $\frac{1}{16}\left(T_{1}+4 T_{1}\right)=30$
or $\sigma_{1}=\frac{30 \times 16}{5}=96$
 Question 2
A schematic of an epicyclic gear train is shown in the figure. The sun (gear 1) and planet (gear 2) are external, and the ring gear (gear 3) is internal. Gear 1, gear 3 and arm OP are pivoted to the ground at O. Gear 2 is carried on the arm OP via the pivot joint at P, and is in mesh with the other two gears. Gear 2 has 20 teeth and gear 3 has 80 teeth. If gear 1 is kept fixed at 0 rpm and gear 3 rotates at 900 rpm counter clockwise (ccw), the magnitude of angular velocity of arm OP is __________rpm (in integer). A 300 B 600 C 900 D 1200
GATE ME 2022 SET-1   Theory of Machine
Question 2 Explanation:
Speed of gear 1 $N_1 =0$,
Speed of gear 3 $N_3 =900rpm$,
Speed of Arm $N_{arm} =?$,
Teeth of gear 1 $z_1=?$
Teeth of gear 2 $z_2=20$
Teeth of gear 3 $z_3=80$
Teeth of gear 2
\begin{aligned} z_3&=z_1+2z_2\\ 80&=z_1+2(20)\\ z_1&=40 \end{aligned}
$\begin{array}{|c|c|c|c|c|c|} \hline \text{S.No}&\text{Condition of mtotion} &\text{Speed of arm} &\text{Speed of gear 1} &\text{Speed of gear 2} &\text{Speed of gear 3}\\ \hline 1&\text{Arm is fixed and gear 1 with +x rev}&0&+1&\frac{-z_1}{z_2}(1)&\frac{-z_1}{z_3}(1)\\ \hline 2&\text{Arm is fixed and gear 1 with +x rev}&0&+x&\frac{-z_1}{z_2}(x)&\frac{-z_1}{z_3}(x)\\ \hline 3&\text{Arm with +y rev}&+y&y&+y&+y\\ \hline 4&\text{Total }&y&x+y &y-x\frac{z_1}{z_2}&y-\frac{z_1}{z_3}(x)\\ \hline \end{array}$
\begin{aligned} N_1&=x+y=0\\ \Rightarrow x&=-y \\ N_3&=y-\frac{z_1}{z_3}(x)=900\\ &=y-\frac{40}{80}(x)=900\\ \Rightarrow y-\frac{40}{80}(-y)=900\\ 1.5y&=900\\ y&=600 rpm \end{aligned}

 Question 3
A power transmission mechanism consists of a belt drive and a gear train as shown in the figure. Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure. The speed and direction of rotation of gear 7, respectively, are
 A 255.68 rpm; clockwise B 255.68 rpm; anticlockwise C 575.28 rpm; clockwise D 575.28 rpm; anticlockwise
GATE ME 2021 SET-2   Theory of Machine
Question 3 Explanation:
\begin{aligned} T_{3} &=44 \\ T_{6} &=36 \\ T_{2} &=18 \\ T_{4} &=15 \\ \frac{N_{1}}{N_{0}} &=\frac{d_{0}}{d_{1}} \\ \Rightarrow\qquad N_{1} &=\frac{150}{250} \times 2500\\ N_{1}&=15,00=N_{2} \\ N_{3}&=\frac{N_{2} \times T_{2}}{T_{3}}=\frac{1500 \times 18}{44}=613.63636=N_{4} \\ N_{6}&=\frac{N_{4} \times T_{4}}{T_{6}}=\frac{613.63636 \times 15}{36} \\ &=255.6818=N_{7}(\text { Clockwise }) \end{aligned}
Gear 5 is idler.
 Question 4
The sun (S) and the planet (P) of an epicyclic gear train shown in the figure have identical number of teeth If the sun (S) and the outer ring (R) gears are rotated in the same direction with angular speed $\omega _S \; and \; \omega _R$, respectively, then the angular speed of the arm AB is
 A $\frac{3}{4}\omega _R + \frac{1}{4}\omega _S$ B $\frac{1}{4}\omega _R + \frac{3}{4}\omega _S$ C $\frac{1}{2}\omega _R - \frac{1}{2}\omega _S$ D $\frac{3}{4}\omega _R - \frac{1}{4}\omega _S$
GATE ME 2020 SET-2   Theory of Machine
Question 4 Explanation:
\begin{aligned} r_{s}+2 r_{p}&=r_{R} \\ \Rightarrow \quad T_{s}+2 T_{p}&=T_{R} \quad (T_{P}=T_{S})\\ 3 T_{P}&=T_{R} \\ \Rightarrow \quad 3 T_{p}&=T_{R} \end{aligned} \begin{aligned} y+x&=\omega_{S} \\ y-\frac{x}{3}&=\omega_{R}\\ & \text{Substract by,}\\ \frac{4 x}{3} &=\left(\omega_{S}-\omega_{R}\right) \\ x &=\frac{3}{4}\left(\omega_{S}-\omega_{R}\right) \\ y &=\omega_{S}-x=\omega_{S}-\frac{3}{4}\left(\omega_{S}-\omega_{R}\right) \\ &=\omega_{S}-\frac{3 \omega_{S}}{4}+\frac{3 \omega_{R}}{4}=\frac{\omega_{S}}{4}+\frac{3 \omega_{R}}{4} \end{aligned}
 Question 5
A balanced rigid disc mounted on a rigid rotor has four identical point masses, each of 10 grams, attached to four points on the 100 mm radius circle shown in the figure. The rotor is driven by a motor at uniform angular speed of 10 rad/s. If one of the masses gets detached then the magnitude of the resultant unbalance force on the rotor is ______ N. (round off to 2 decimal places).
 A 0.1 B 1 C 10 D 0.001
GATE ME 2020 SET-1   Theory of Machine
Question 5 Explanation: $\omega=10 \mathrm{rad} / \mathrm{s}, r=100 \mathrm{mm}=0.1 \mathrm{m}$
If one mass is detached then Now, unbalance tone, F=$m r \omega^{2}$
$\begin{array}{l} =\frac{10}{1000} \times 0.1 \times(10)^{2}=\frac{10 \times 0.1 \times 100}{1000} \\ =0.1 \mathrm{N} \end{array}$

There are 5 questions to complete.

### 2 thoughts on “Gear and Gear Train”

1. Sir,I think 19ans wrong ans:10rpm anticlockwise direction

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