# Gear and Gear Train

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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=$m r \omega^{2}$
$\begin{array}{l} =\frac{10}{1000} \times 0.1 \times(10)^{2}=\frac{10 \times 0.1 \times 100}{1000} \\ =0.1 \mathrm{N} \end{array}$
 Question 5
A spur gear has pitch circle diameter D and number of teeth T. The circular pitch of the gear is
 A $\frac{\pi D}{T}$ B $\frac{T}{D}$ C $\frac{D}{T}$ D $\frac{2 \pi D}{T}$
GATE ME 2019 SET-2   Theory of Machine
Question 5 Explanation:
Circular pitch : It is the distance between two similar points on adjacent teeth measured along pitch
circle circumference circular pitch
$\left(P_{c}\right)=\frac{\text { Pitch circlecircum }}{\text { Number of teeth }}=\frac{\pi D}{T}$
 Question 6
A spur gear with 20$^{\circ}$ full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
 A 0.36 kN B 0.73 kN C 1.39 kN D 2.78kN
GATE ME 2019 SET-1   Theory of Machine
Question 6 Explanation:
$\begin{array}{l} \phi=20^{\circ}, P=20 k W, \omega=200 \mathrm{rad} / \mathrm{s}, d=100 \mathrm{mm}=0.1 \mathrm{m} \\ \text { Torque }=\text { Power } / \omega \\ \mathrm{T}=\frac{20000}{200}=100 \mathrm{Nm} \\ \text { Now, } \mathrm{T}=\mathrm{F}_{\mathrm{T}} \times \frac{\mathrm{d}}{2} \\ \Rightarrow 100=\mathrm{F}_{\mathrm{T}} \times \frac{0.1}{2} \\ \Rightarrow \mathrm{F}_{\mathrm{T}}=2000 \mathrm{N}\\ \frac{F_{R}}{F_{T}}=\tan \phi \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=2000 \times \tan 20^{\circ} \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=727.94 \mathrm{N}=0.73 \mathrm{kN} \end{array}$
 Question 7
A frictionless gear train is shown in the figure. The leftmost 12-teeth gear is given a torque of 100 N-m. The output torque from the 60-teeth gear on the right in N-m is A 5 B 20 C 500 D 2000
GATE ME 2018 SET-2   Theory of Machine
Question 7 Explanation: \begin{aligned} \tau_{1}&=100 \mathrm{Nm}\\ \text{Let speed of 1 is }N_{1} \\ (1,2):\qquad N_{2}&=N_{1} \times \frac{T_{1}}{T_{2}}=N_{1} \times \frac{12}{48}=\frac{N_{1}}{4} \\ N_{3}=N_{2}&=\frac{N_{1}}{4}\\ (3,4) \qquad N_{4}&=N_{3} \times \frac{T_{3}}{T_{4}}=\frac{N_{1}}{4} \times \frac{12}{60} \\ N_{4}&=\frac{N_{1}}{20} \end{aligned}
By Power conservation
(Assume $\eta$ (efficiency) =1)
\begin{aligned} \tau_{1} \times N_{1} &=\tau_{4} \times N_{4} \\ 100 \times N_{1} &=\tau_{4} \times \frac{N_{1}}{20} \\ \tau_{4} &=2000 \mathrm{N}-\mathrm{m} \end{aligned}
 Question 8
An epicyclic gear train is shown in the figure below. The number of teeth on the gears A, B and D are 20, 30 and 20, respectively. Gear C has 80 teeth on the inner surface and 100 teeth on the outer surface. If the carrier arm AB is fixed and the sun gear A rotates at 300 rpm in the clockwise direction, then the rpm of D in the clockwise direction is A 240 B -240 C 375 D -375
GATE ME 2018 SET-1   Theory of Machine
Question 8 Explanation:
$T_{A}=20, T_{B}=30, T_{D}=20, T_{C}=80(\text { Inner }), T_{C}=100 \text { (Outer) }$
Arm is fixed, no epicyclic nature. Taking clockwise direction as positive.
\begin{aligned} N_{A}&=+300 \\ (A, B) \qquad N_{B}&=-\frac{300 \times 20}{30}=-200\\ (B, C)\qquad N_{C}&=-200 \times \frac{30}{80}=-75\\ (C, D)\qquad N_{D}&=+75 \times \frac{100}{20}=+375 \end{aligned}
 Question 9
A gear train shown in the figure consists of gear P, Q, R and S. Gear Q and gear R are mounted on the same shaft. All the gears are mounted on parallel shafts and the number of teeth of P, Q, R and S are 24, 45, 30 and 80, respectively. Gear P is rotating at 400 rpm. The speed (in rpm) of the gear S is _____ A 90 B 120 C 245 D 150
GATE ME 2017 SET-2   Theory of Machine
Question 9 Explanation:
$\begin{array}{ll} T_{P}=24 \\ T_{Q}=45 & N_{P}=400 \mathrm{rpm} \\ T_{R}=30 & N_{S}=? \\ T_{S}=80 \end{array}$
$Here gear R is not meshing at all.
[latex] \begin{array}{l} \frac{N_{P}}{N_{Q}}=\frac{T_{O}}{T_{P}}=\frac{45}{24} \qquad \cdots(1)\\ \frac{N_{Q}}{N_{S}}=\frac{T_{S}}{T_{Q}}=\frac{80}{45}\qquad \cdots(2)\\ \text{By }(1) \times(2) \\ \frac{N_{P}}{N_{S}}=\frac{45}{24} \times \frac{80}{45} \\ N_{S}=N_{P} \times \frac{24}{80}=\frac{400 \times 24}{80} \\ N_{S}=120 \text { r.p.m } \end{array}$
 Question 10
In an epicyclic gear train, shown in the figure, the outer ring gear is fixed, while the sun gear rotates counterclockwise at 100rpm. Let the number of teeth on the sun, planet and outer gears to be 50, 25, and 100, respectively. The ratio of magnitudes of angular velocity of the planet gear to the angular velocity of the carrier arm is _________. A 3 B 4 C 5 D 6
GATE ME 2017 SET-1   Theory of Machine
Question 10 Explanation:
We take convention clockwise (+ve)
\begin{aligned} N_{D} &=0 \\ N_{S} &=-100 \\ T_{S} &=50, \quad T_{P}=25, T_{D}=100 \\ \frac{N_{P}}{N_{\mathrm{ARM}}}&=? \end{aligned} $y+x=-100 \qquad \cdots(i)$
$y-\frac{x}{2}=0 \qquad \cdots(ii)$
By equation (i) - (ii)
\begin{aligned} \frac{3 x}{2} &=-100=x=\frac{-200}{3} \\ y &=-100-x=-100+\frac{200}{3}=\frac{-100}{3} \\ N_{p} &=y-2 x=\frac{-100}{3}+\frac{400}{3}=\frac{300}{3}=100 \\ N_{\mathrm{ARM}} &=y=\frac{-100}{3} \\ \frac{N_{P}}{N_{\mathrm{ARM}}} &=\frac{100}{\frac{-100}{3}}=-3 \\ \left| \frac{N_{\mathrm{P}}}{N_{\mathrm{ARM}}}\right | &=3 \end{aligned}
There are 10 questions to complete.

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

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

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