Tooth Numbers for the the Gear Train Shown in the Figure UPDATED
Tooth Numbers for the the Gear Train Shown in the Figure
A power manual mechanism consists of a belt drive and a gear train as shown in the effigy.
Diameters of pulleys of chugalug drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure. The speed and management of rotation of gear 7, respectively, are
| 255.68 rpm; clockwise | |
| 255.68 rpm; anticlockwise | |
| 575.28 rpm; clockwise | |
| 575.28 rpm; anticlockwise |
GATE ME 2021 Set-2 Theory of Machine
Question ane Explanation:
\begin{aligned} T_{three} &=44 \\ T_{6} &=36 \\ T_{2} &=xviii \\ T_{4} &=15 \\ \frac{N_{1}}{N_{0}} &=\frac{d_{0}}{d_{1}} \\ \Rightarrow\qquad N_{one} &=\frac{150}{250} \times 2500\\ N_{1}&=15,00=N_{2} \\ N_{3}&=\frac{N_{2} \times T_{2}}{T_{iii}}=\frac{1500 \times 18}{44}=613.63636=N_{4} \\ N_{6}&=\frac{N_{4} \times T_{iv}}{T_{6}}=\frac{613.63636 \times 15}{36} \\ &=255.6818=N_{7}(\text { Clockwise }) \end{aligned}
Gear 5 is idler.
The sun (S) and the planet (P) of an epicyclic gear railroad train shown in the figure have identical number of teeth
If the sun (S) and the outer band (R) gears are rotated in the same direction with athwart speed \omega _S \; and \; \omega _R, respectively, and so the athwart speed of the arm AB is
| \frac{3}{4}\omega _R + \frac{ane}{4}\omega _S | |
| \frac{1}{iv}\omega _R + \frac{3}{iv}\omega _S | |
| \frac{ane}{two}\omega _R - \frac{1}{2}\omega _S | |
| \frac{three}{four}\omega _R - \frac{1}{iv}\omega _S |
GATE ME 2020 SET-ii Theory of Machine
Question 2 Explanation:
\brainstorm{aligned} r_{s}+2 r_{p}&=r_{R} \\ \Rightarrow \quad T_{s}+two T_{p}&=T_{R} \quad (T_{P}=T_{South})\\ iii T_{P}&=T_{R} \\ \Rightarrow \quad 3 T_{p}&=T_{R} \end{aligned}
\begin{aligned} y+x&=\omega_{Southward} \\ y-\frac{x}{3}&=\omega_{R}\\ & \text{Substract by,}\\ \frac{4 10}{three} &=\left(\omega_{Due south}-\omega_{R}\right) \\ 10 &=\frac{3}{four}\left(\omega_{S}-\omega_{R}\right) \\ y &=\omega_{S}-x=\omega_{S}-\frac{three}{4}\left(\omega_{South}-\omega_{R}\right) \\ &=\omega_{S}-\frac{3 \omega_{S}}{iv}+\frac{iii \omega_{R}}{iv}=\frac{\omega_{Southward}}{4}+\frac{iii \omega_{R}}{4} \end{aligned}
A balanced rigid disc mounted on a rigid rotor has four identical indicate masses, each of ten grams, attached to iv points on the 100 mm radius circle shown in the figure.
The rotor is driven by a motor at uniform athwart speed of x rad/southward. If one of the masses gets detached so the magnitude of the resultant unbalance strength on the rotor is ______ N. (round off to 2 decimal places).
| 0.one | |
| 1 | |
| 10 | |
| 0.001 |
GATE ME 2020 Set up-1 Theory of Automobile
Question 3 Explanation:
\omega=10 \mathrm{rad} / \mathrm{s}, r=100 \mathrm{mm}=0.1 \mathrm{g}
If one mass is discrete then
At present, unbalance tone, F=m r \omega^{2}
\brainstorm{array}{l} =\frac{10}{1000} \times 0.1 \times(x)^{2}=\frac{10 \times 0.1 \times 100}{1000} \\ =0.ane \mathrm{N} \end{array}
A spur gear has pitch circle diameter D and number of teeth T. The circular pitch of the gear is
| \frac{\pi D}{T} | |
| \frac{T}{D} | |
| \frac{D}{T} | |
| \frac{2 \pi D}{T} |
GATE ME 2019 SET-ii Theory of Machine
Question iv Explanation:
Circular pitch : Information technology is the altitude between two like points on adjacent teeth measured along pitch
circle circumference round pitch
\left(P_{c}\right)=\frac{\text { Pitch circlecircum }}{\text { Number of teeth }}=\frac{\pi D}{T}
A spur gear with 20^{\circ} full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circumvolve diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
| 0.36 kN | |
| 0.73 kN | |
| 1.39 kN | |
| 2.78kN |
GATE ME 2019 SET-one Theory of Machine
Question 5 Explanation:
\begin{array}{fifty} \phi=20^{\circ}, P=twenty yard Westward, \omega=200 \mathrm{rad} / \mathrm{s}, d=100 \mathrm{mm}=0.i \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}
A frictionless gear train is shown in the figure. The leftmost 12-teeth gear is given a torque of 100 Due north-chiliad. The output torque from the 60-teeth gear on the right in North-g is
| 5 | |
| 20 | |
| 500 | |
| 2000 |
GATE ME 2018 SET-2 Theory of Machine
Question half-dozen Caption:
\begin{aligned} \tau_{1}&=100 \mathrm{Nm}\\ \text{Allow speed of 1 is }N_{1} \\ (1,two):\qquad N_{ii}&=N_{1} \times \frac{T_{one}}{T_{two}}=N_{1} \times \frac{12}{48}=\frac{N_{one}}{iv} \\ N_{iii}=N_{2}&=\frac{N_{ane}}{four}\\ (3,4) \qquad N_{4}&=N_{iii} \times \frac{T_{three}}{T_{four}}=\frac{N_{1}}{4} \times \frac{12}{60} \\ N_{iv}&=\frac{N_{one}}{twenty} \end{aligned}
By Power conservation
(Assume \eta (efficiency) =1)
\begin{aligned} \tau_{1} \times N_{ane} &=\tau_{4} \times N_{iv} \\ 100 \times N_{1} &=\tau_{4} \times \frac{N_{1}}{xx} \\ \tau_{4} &=2000 \mathrm{N}-\mathrm{m} \terminate{aligned}
An epicyclic gear train is shown in the figure below. The number of teeth on the gears A, B and D are twenty, thirty and twenty, 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 management, then the rpm of D in the clockwise direction is
| 240 | |
| -240 | |
| 375 | |
| -375 |
GATE ME 2018 SET-1 Theory of Auto
Question vii Caption:
T_{A}=20, T_{B}=30, T_{D}=20, T_{C}=80(\text { Inner }), T_{C}=100 \text { (Outer) }
Arm is stock-still, no epicyclic nature. Taking clockwise direction every bit 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{xxx}{80}=-75\\ (C, D)\qquad N_{D}&=+75 \times \frac{100}{20}=+375 \end{aligned}
A gear train shown in the figure consists of gear P, Q, R and South. 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 Due south is _____
| xc | |
| 120 | |
| 245 | |
| 150 |
GATE ME 2017 Set up-two Theory of Automobile
Question 8 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.<br>[latex] \begin{array}{50} \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{lxxx}{45}\qquad \cdots(two)\\ \text{By }(1) \times(two) \\ \frac{N_{P}}{N_{Due south}}=\frac{45}{24} \times \frac{lxxx}{45} \\ N_{S}=N_{P} \times \frac{24}{80}=\frac{400 \times 24}{80} \\ N_{Southward}=120 \text { r.p.g } \end{array}
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 fifty, 25, and 100, respectively. The ratio of magnitudes of angular velocity of the planet gear to the angular velocity of the carrier arm is _________.
| 3 | |
| 4 | |
| 5 | |
| 6 |
GATE ME 2017 Fix-1 Theory of Machine
Question 9 Explanation:
We take convention clockwise (+ve)
\brainstorm{aligned} N_{D} &=0 \\ N_{S} &=-100 \\ T_{Due south} &=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{three x}{ii} &=-100=x=\frac{-200}{3} \\ y &=-100-x=-100+\frac{200}{iii}=\frac{-100}{3} \\ N_{p} &=y-two ten=\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 | &=iii \end{aligned}
In the gear train shown, gear 3 is carried on arm 5. Gear 3 meshes with gear two and gear four. The number of teeth on gear two, 3, and iv are 60, twenty, and 100, respectively. If gear 2 is fixed and gear 4 rotates with an angular velocity of 100 rpm in the counterclockwise direction, the angular speed of arm 5 (in rpm) is
| 166.7 counterclockwise | |
| 166.seven clockwise | |
| 62.5 counterclockwise | |
| 62.v clockwise |
GATE ME 2016 Fix-1 Theory of Machine
Question 10 Explanation:
Given : Gear two is fixed
\therefore \quad N+y=0 \qquad [\therefore N=-y]
Gear 4 rotates 100 rpm is counter clockwise.
\begin{aligned} -\frac{N \times 60}{100}+y &=-100 \\ +y \times 0.6+y &=-100 \\ y &=\frac{-100}{i.6}=-62.5 \terminate{aligned}
Hence angular speed of arm is 62.5
(counterclockwise).
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Tooth Numbers for the the Gear Train Shown in the Figure UPDATED
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