Question 1 |
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
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
255.68 rpm; clockwise | |
255.68 rpm; anticlockwise | |
575.28 rpm; clockwise | |
575.28 rpm; anticlockwise |
Question 1 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.
Gear 5 is idler.
Question 2 |
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
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
\frac{3}{4}\omega _R + \frac{1}{4}\omega _S | |
\frac{1}{4}\omega _R + \frac{3}{4}\omega _S | |
\frac{1}{2}\omega _R - \frac{1}{2}\omega _S | |
\frac{3}{4}\omega _R - \frac{1}{4}\omega _S |
Question 2 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}
\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 3 |
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).
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).
0.1 | |
1 | |
10 | |
0.001 |
Question 3 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 4 |
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} |
Question 4 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}
circle circumference circular pitch
\left(P_{c}\right)=\frac{\text { Pitch circlecircum }}{\text { Number of teeth }}=\frac{\pi D}{T}
Question 5 |
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
0.36 kN | |
0.73 kN | |
1.39 kN | |
2.78kN |
Question 5 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 6 |
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
5 | |
20 | |
500 | |
2000 |
Question 6 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 7 |
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
240 | |
-240 | |
375 | |
-375 |
Question 7 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}
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 8 |
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 _____
90 | |
120 | |
245 | |
150 |
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}{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}
Here gear R is not meshing at all.<br>[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 9 |
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 _________.
3 | |
4 | |
5 | |
6 |
Question 9 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}
\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}
Question 10 |
In the gear train shown, gear 3 is carried on arm 5. Gear 3 meshes with gear 2 and gear 4. The number of teeth on gear 2, 3, and 4 are 60, 20, 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 | |
166.7 clockwise | |
62.5 counterclockwise | |
62.5 clockwise |
Question 10 Explanation:
Given : Gear 2 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}{1.6}=-62.5 \end{aligned}
Hence angular speed of arm is 62.5
(counterclockwise).
There are 10 questions to complete.
Sir,I think 19ans wrong ans:10rpm anticlockwise direction
Ans is 10rpm ccw (Counter clockwise)