# Theory of Machine

 Question 1
The torque provided by an engine is given by $T(\theta ) = 12000 + 2500 sin(2\theta ) N.m$, where $\theta$ is the angle turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a machine that provides a constant resisting torque. If variation of the speed from the mean speed is not to exceed $\pm 0.5%$, the minimum mass moment of inertia of the flywheel should be _______ $kg.m^2$ (round off to the nearest integer).
 A 245 B 360 C 570 D 640
GATE ME 2021 SET-2      Flywheel
Question 1 Explanation:

\begin{aligned} \omega &=\frac{\pi \times 200}{30}=20.9439 \mathrm{rad} / \mathrm{s} \\ \Delta E &=\int_{0}^{\frac{\pi}{2}}\left(T-T_{\text {mean }}\right) d \theta=2500 \int_{0}^{\frac{\pi}{2}} \sin 2 \theta d \theta \\ &=2500 \times 1=2500 \mathrm{~J} \\ \Delta E &=I \omega^{2} C_{s} \\ 2500 &=I \times 20.9439^{2} \times 0.01 \\ I &=569.934 \mathrm{kgm}^{2} \simeq 570 \mathrm{~kg} . \mathrm{m}^{2} \end{aligned}
 Question 2
The wheels and axle system lying on a rough surface is shown in the figure.

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the moment of inertia of the axle and assume $g=9.8 m/s^2$. An effort of 10 N is applied on the axle in the horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface without slip. The acceleration of the wheel axle system in horizontal direction is ________$m/s^2$(round off to one decimal place).
 A 2.8 B 3.6 C 5 D 6.4
GATE ME 2021 SET-2      Displacement, Velocity and Acceleration
Question 2 Explanation:

\begin{aligned} I_G&=2 \times (M \times r^2) \\&=2 \times 1 \times 0.4^2=0.32\; kg/m^2 \\ \Sigma F&=m\cdot a\\ &\Rightarrow 10-f=3.5 \times a\;...(i)\\ \Sigma T&=I\cdot \alpha \\ &\Rightarrow 10 \times 0.1 -f\times 0.4=0.32 \times \frac{a}{0.4} \;...(ii)\\ \end{aligned}
[As there is no slip $\therefore \;\;a=r\alpha$ ]
Solving (i) and (ii),
$\therefore \;\;a=5\; m/s^2$
 Question 3
Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.

The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position. The natural frequency of the system is
 A $\sqrt{\frac{kr^2}{J-mr^2}}$ B $\sqrt{\frac{kr^2}{J+mr^2}}$ C $\sqrt{\frac{k}{m}}$ D $\sqrt{\frac{kr^2}{J}}$
GATE ME 2021 SET-2      Vibration
Question 3 Explanation:

\begin{aligned} E &=\frac{1}{2} m \dot{x}^{2}+\frac{1}{2} k x^{2}+\frac{1}{2} I \omega^{2} \\ &=\frac{1}{2} m r^{2} \dot{\theta}^{2}+\frac{1}{2} k r^{2} \theta^{2}+\frac{1}{2} J \dot{\theta}^{2} \\ &=\frac{1}{2}\left[\left(J+m r^{2}\right) \dot{\theta}^{2}+k r^{2} \theta^{2}\right]=0 \\ \frac{d E}{d t} &=0 \\ \left(J+m r^{2}\right) \times 2 \dot{\theta} \ddot{\theta}+k r^{2} 2 \theta \dot{\theta} &=0 \\ \left(\ddot{\theta}+\frac{k r^{2}}{J+m r^{2}}\right) \theta &=0 \\ \omega_{n} &=\sqrt{\frac{k r^{2}}{J+m r^{2}}} \end{aligned}
 Question 4
A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The isolator can be considered as a combination of stiffness (K) and damper (damping factor, $\xi$) in parallel. The designer has the following four isolators:

1) K = 640 kN/m, $\xi$ = 0.70
2) K = 640 kN/m, $\xi$ = 0.07
3) K = 22.5 kN/m, $\xi$ = 0.70
4) K = 22.5 kN/m, $\xi$ = 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.
 A 1-3-4-2 B 1-3-2-4 C 4-3-1-2 D 3-1-2-4
GATE ME 2021 SET-2      Vibration
Question 4 Explanation:
Transmitted force problem:
\begin{aligned} \omega&=40 \mathrm{rad} / \mathrm{s} \text { (force frequency) }\\ m&=100 \mathrm{~kg} \\ \epsilon&=\frac{\sqrt{1+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}}}{\sqrt{\left\{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}+\left\{\frac{2 \xi \omega}{\omega_{n}}\right\}^{2}\right\}}} \\ \omega_{n}&=\sqrt{\frac{640 \times 10^{3}}{100}}=80 \Rightarrow \frac{\omega}{\omega_{n}}=0.5\\ \omega_{n}&=\sqrt{\frac{22.5 \times 10^{3}}{100}}=47.434164 \Rightarrow \frac{\omega}{\omega_{n}}=\frac{40}{15}=2.666 \end{aligned}

 Question 5
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      Gear and Gear Train
Question 5 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 6
The controlling force curves P, Q and R for a spring controlled governor are shown in the figure, where $r_1$ and $r_2$ are any two radii of rotation.

The characteristics shown by the curves are
 A P - Unstable; Q - Stable; R - Isochronous B P - Unstable; Q - Isochronous; R - Stable C P- Stable; Q - Isochronous; R - Unstable D P - Stable; Q - Unstable; R - Isochronous
GATE ME 2021 SET-2      Flywheel
Question 6 Explanation:
$F(r)|_P=ar+b \quad \rightarrow \text{Unstable}$
$F(r)|_Q=ar+b \quad \rightarrow \text{Isochronous}$
$F(r)|_R=ar-b \quad \rightarrow \text{Stable}$
 Question 7
Consider the mechanism shown in the figure. There is rolling contact without slip between the disc and ground.

Select the correct statement about instantaneous centers in the mechanism.
 A Only points P, Q, and S are instantaneous centers of mechanism B Only points P, Q, S and T are instantaneous centers of mechanism C Only points P, Q, R, S, and U are instantaneous centers of mechanism D All points P, Q, R, S, T and U are instantaneous centers of mechanism
GATE ME 2021 SET-2      Displacement, Velocity and Acceleration
Question 7 Explanation:

Points P, Q, R, S, T and U are instantaneous centers of mechanism.
 Question 8
The Whitworth quick return mechanism is shown in the figure with link lengths as follows: OP = 300 mm, OA = 150 mm, AR = 160 mm, RS = 450 mm.

The quick return ratio for the mechanism is ________(round off to one decimal place).
 A 1 B 2 C 2.5 D 3.5
GATE ME 2021 SET-1      Dynamic Analysis of Slider-crank
Question 8 Explanation:

\begin{aligned} \cos \frac{\alpha}{2} &=\frac{150}{300}=\frac{1}{2} \\ \Rightarrow \qquad \frac{\alpha}{2} &=60^{\circ} \\ \alpha &=120^{\circ} \\ \beta &=\left(360^{\circ}-120^{\circ}\right)=240^{\circ} \\ \mathrm{QRR} &=\left(\frac{\beta}{\alpha}\right)=\frac{240}{120}=2 \end{aligned}
 Question 9
A tappet valve mechanism in an IC engine comprises a rocker arm ABC that is hinged at B as shown in the figure. The rocker is assumed rigid and it oscillates about the hinge B. The mass moment of inertia of the rocker about B is $10^{-4}\;kg.m^2$. The rocker arm dimensions are a = 3.5 cm and b = 2.5 cm. A pushrod pushes the rocker at location A, when moved vertically by a cam that rotates at N rpm. The pushrod is assumed massless and has a stiffness of 15 N/mm. At the other end C, the rocker pushes a valve against a spring of stiffness 10 N/mm. The valve is assumed massless and rigid.

Resonance in the rocker system occurs when the cam shaft runs at a speed of ______rpm (round off to the nearest integer).
 A 496 B 4739 C 790 D 2369
GATE ME 2021 SET-1      Vibration
Question 9 Explanation:

$I=10^{-4} \mathrm{~kg}-\mathrm{m}^{2}$
By D'Alembert Principle
$I \ddot{\theta}+\left[10000 \times(0.025)^{2}+15000 \times(0.035)^{2}\right] \theta=0$
\begin{aligned} \left(10^{-4}\right) \ddot{\theta}+(24.625) \theta&=0 \\ \ddot{\theta}+\left(\frac{24.625}{10^{-4}}\right) \theta&=0\\ \Rightarrow \qquad\qquad \omega_{n}^{2}&=(246250) \\ \omega_{n}&= 496.2358\; \text{rad/s} \\ \Rightarrow \qquad\qquad N_C&=\frac{496.2358 \times 60}{2 \pi} \\ &=4738.70 \text{ rpm}\end{aligned}

 Question 10
Consider a two degree of freedom system as shown in the figure, where PQ is a rigid uniform rod of length, b and mass, m.

Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is
 A $\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta$ B $\frac{1}{2}\left ( M+m \right )\dot x^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot\theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta$ C $\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2}$ D $\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}mb\dot{\theta }\dot{x}\cos\theta +\frac{1}{6}mb^{2}\dot \theta ^{2}-\frac{1}{2}kx^{2}+mg\frac{b}{2}\cos\theta +Fb\sin\theta$
GATE ME 2021 SET-1      Vibration
Question 10 Explanation:

\begin{aligned} \text { For mass, } T&=\frac{1}{2} M \dot{x}^{2}\\ V&=\frac{1}{2} K x^{2} \end{aligned}
For mass, m

\begin{aligned} d m &=\frac{m}{b} d y \\ \text { Displacement } &=x+y \sin \theta \\ \text { Velocity } &=\dot{x}+y \cos \theta \dot{\theta} \\ d T & =\frac{1}{2} d M V e l^{2}=\frac{1}{2} d m\left[\dot{x}^{2}+y^{2} \dot{\theta}^{2} \cos ^{2} \theta+2 \dot{x} \dot{\theta} y \cos \theta\right] \\ T &=d T=\frac{1}{2} \frac{m}{b} \int\left(\dot{x}^{2}+y^{2} \dot{\theta}^{2} \times 1+2 \dot{x} \dot{\theta} y \cos \theta\right) d y \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+2 \dot{x} \dot{\theta} \cos \theta \frac{b^{2}}{2}\right) \\ T &=\frac{1}{2} \frac{m}{b}\left(\dot{x}^{2} b+\frac{b^{3}}{3} \dot{\theta}^{2}+\dot{x} \dot{\theta} b^{2} \cos \theta\right) \\ T &=\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} \dot{x} \dot{\theta} b \cos \theta \\ V &=-m g \frac{b}{2} \cos \theta \end{aligned} For both masses, M and m
\begin{aligned} T &=\frac{1}{2} M \dot{x}^{2}+\frac{1}{2} m \dot{x}^{2}+\frac{m b^{2}}{6} \dot{\theta}^{2}+\frac{m}{2} b \dot{\theta} \dot{x}^{2} \cos \theta \\ V &=\frac{1}{2} k x^{2}-m g \frac{b}{2} \cos \theta \\ L &=T-V \\ &=\frac{1}{2}(M+m) \dot{x}^{2}+\frac{1}{2} m b \dot{x} \dot{c} \cos \theta+\frac{m b^{2} \dot{\theta}^{2}}{6}-\frac{1}{2} k x^{2}+m g \frac{b}{2} \cos \theta \end{aligned}

There are 10 questions to complete.

### 8 thoughts on “Theory of Machine”

1. figures of questions are missing,and in some question statement is given wrong..

• Thank you prakash for your suggestions.
Can you please share the questions numbers for the above suggestions.

• We have review the question 8 with original published paper by IIT. There is only one figure which is there in our question -8. It might possible due to slow internet image takes time to load.