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)


52 | |
48 | |
96 | |
112 |
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).


300 | |
600 | |
900 | |
1200 |
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}
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

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 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.
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

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 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}

\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).

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 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.
Sir,I think 19ans wrong ans:10rpm anticlockwise direction
Ans is 10rpm ccw (Counter clockwise)