Theory of Machine

Question 1
In the configuration of the planar four-bar mechanism at a certain instant as shown in the figure, the angular velocity of the 2 cm long link is \omega _2 =5rad/s. Given the dimensions as shown, the magnitude of the angular velocity \omega _4 of the 4 cm long link is given by _____ rad/s (round off to 2 decimal places).

A
1.25
B
0.75
C
2.25
D
3.75
GATE ME 2022 SET-2      Planar Mechanisms
Question 1 Explanation: 


Number of links = 4
Number of I-centers 4c_2=6 \omega _2=5 \; rad/sec
\omega _4=?
For
\begin{aligned} V_{I_{24}}&=I_{12}I_{24}\omega _{2}=I_{14}I_{24}\omega_4\\ &=2 \times 5 =8 \times \omega_4\\ \omega_4&=1.25 \; rad/sec \end{aligned}
Question 2
A spring mass damper system (mass m, stiffness k, and damping coefficient c) excited by a force F(t)=B\sin \omega t , where B,\omega, t are the amplitude, frequency and time, respectively, is shown in the figure. Four different responses of the system (marked as (i) to (iv)) are shown just to the right of the system figure. In the figures of the responses, A is the amplitude of response shown in red color and the dashed lines indicate its envelope. The responses represent only the qualitative trend and those are not drawn to any specific scale.

Four different parameter and forcing conditions are mentioned below.

(P)\;\;C \gt 0 \text{ and }\omega =\sqrt{k/m}
(Q)\;\;C \lt 0 \text{ and }\omega \neq 0
(R)\;\;C=0 \text{ and }\omega =\sqrt{k/m}
(S)\;\;C=0 \text{ and }\omega \cong \sqrt{k/m}

Which one of the following options gives correct match (indicated by arrow \rightarrow ) of the parameter and forcing conditions to the responses?
A
(P) \rightarrow (i), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (ii)
B
(P) \rightarrow (ii), (Q) \rightarrow (iii), (R) \rightarrow (iv), (S) \rightarrow (i)
C
(P) \rightarrow (i), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (iii)
D
(P) \rightarrow (iii), (Q) \rightarrow (iv), (R) \rightarrow (ii), (S) \rightarrow (i)
GATE ME 2022 SET-2      Vibration
Question 2 Explanation: 
FBD of mass

m\ddot{x}+c\dot{x}+kx=F(t)
Solution of differential equation
x(t) = (C.F) + (P.I)
Considering condition (P)
If C > 0 and \omega =\sqrt{k/m}
For this condition the displacement (x) is given by
[Assume the system to be under damped]
x(t)=X_0e^{-\zeta \omega _nt} \cos (\omega _dt -\phi )+ X \cos (\omega t-\phi )
Transient response + steady state response
As t\rightarrow \infty the transient response decays to zero and only steady state response will remain
x(t)=X \cos (\omega t-\phi )
For this condition the response curve will be


Considering condition (Q)
c \gt 0 \text{ and } \omega \neq 0
The differential equation becomes
m\ddot{x}-c\dot{x}+kx=F(t)
Solution of above differential equation is
x(t)=X_0e^{+c_1t} \cos (\omega _dt -\phi )+ X \cos (\omega t-\phi )
As t\rightarrow \infty the transient response approaches to \infty and increases exponentially
The plot will be


Considering condition (R)
C=0,\omega =\sqrt{\frac{k}{m}} \;\;(Resonance)
The differential equation is
m\ddot{x}+kx=F(t)
Solution for above differential equation
x(t)=x_o \cos \omega _nt+\frac{\dot{x_0}}{\omega _n} \sin \omega _n t+\underbrace{\frac{x_{static}\omega _nt}{2} \sin (\omega _n t)}_{\text{Incraeses with time linearly}}
So the correct plot will be


Considering condition (S)
C=0,\omega \cong \sqrt{\frac{k}{m}}
If the force frequency is close to, but not exactly equal to, natural frequency of the system, a phenomenon is known as beating. In this kind of vibration the amplitude buils up and then diminishes in a regular pattern.
The displacement can be expressed as
x(t)=\left ( \frac{B}{m} \right )\left [ \frac{\cos \omega t- \cos \omega _n t}{\omega _n^2-\omega } \right ]
The correct plot will be

Question 3
A rigid body in the X-Y plane consists of two point masses (1 kg each) attached to the ends of two massless rods, each of 1 cm length, as shown in the figure. It rotates at 30 RPM counter-clockwise about the Z-axis passing through point O. A point mass of \sqrt{2} kg, attached to one end of a third massless rod, is used for balancing the body by attaching the free end of the rod to point O. The length of the third rod is ________ cm.

A
1
B
\sqrt{2}
C
\frac{1}{\sqrt{2}}
D
\frac{1}{2\sqrt{2}}
GATE ME 2022 SET-2      Balancing
Question 3 Explanation: 


m_1=1kg, m_2=1kg, r_1=1cm, r_2=1cm
Balancing mass, m_b=\sqrt{2}kg
Making force polygon for complete balance

From the right angle triangle,
\begin{aligned} (m_br_b)^2&=(m_1r_1)^2+(m_2r_2)^2 \\ (\sqrt{2}r_b)^2&=(1\times 1)^2+(1\times 1)^2 \\ 2r_b^2&=2 \\ r_b&=1 \end{aligned}
Question 4
For a dynamical system governed by the equation,
\ddot{x}(t)+2\varsigma \omega _n\dot{x}(t)+\omega _n^2x(t)=0
the damping ratio \varsigma is equal to \frac{1}{2\pi} \log_e 2 . The displacement x of this system is measured during a hammer test. A displacement peak in the positive displacement direction is measured to be 4 mm. Neglecting higher powers ( > 1) of the damping ratio, the displacement at the next peak in the positive direction will be _______ mm (in integer)
A
1
B
2
C
3
D
4
GATE ME 2022 SET-2      Vibration
Question 4 Explanation: 
For a dynamic system,
\ddot{x}+2\zeta \omega _n\dot{x}+\omega _n^2x=0
The above equation represents a damped free vibration system.
Given damping factor, \zeta=\frac{1}{2 \pi} \ln (2)
Given a peak x_{n-1}=4 \; mm
x_n=?
Logarithmic decrement is given by
\ln\left ( \frac{x_{n-1}}{x_n} \right )=\frac{2 \pi\zeta }{\sqrt{1-\zeta ^2}}
As question is asking to neglect the higher power of \zeta , so neglecting \zeta ^2 , as \zeta \lt 1 .
\begin{aligned} \ln \left ( \frac{x_{n-1}}{x_n} \right )&=2 \pi\zeta \\ 2 \pi\zeta &= \ln (2) \\ \Rightarrow \ln \left ( \frac{x_{n-1}}{x_n} \right ) &= \ln (2)\\ \Rightarrow \frac{4}{x_n}&=2 \\ \Rightarrow x_n&=2\; mm \end{aligned}
Question 5
A massive uniform rigid circular disc is mounted on a frictionless bearing at the end E of a massive uniform rigid shaft AE which is suspended horizontally in a uniform gravitational field by two identical light inextensible strings AB and CD as shown, where G is the center of mass of the shaft-disc assembly and g is the acceleration due to gravity. The disc is then given a rapid spin \omega about its axis in the positive x-axis direction as shown, while the shaft remains at rest. The direction of rotation is defined by using the right-hand thumb rule. If the string AB is suddenly cut, assuming negligible energy dissipation, the shaft AE will

A
rotate slowly (compared to \omega ) about the negative z-axis direction
B
rotate slowly (compared to \omega ) about the positive z-axis direction
C
rotate slowly (compared to \omega ) about the negative y-axis direction
D
rotate slowly (compared to \omega ) about the positive y-axis direction
GATE ME 2022 SET-2      Gyroscope
Question 5 Explanation: 


The spin vector will chase the couple on torque vector and produce precision in system.
Hence precision will be -y direction. Rotate slowly (compared to \omega ) about negative z-axis direction.
Question 6
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      Gear and Gear Train
Question 6 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 7
Consider a forced single degree-of-freedom system governed by
\ddot{x}(t)+2\zeta \omega _n\dot{x}(t)+\omega _n^2x(t)=\omega _n^2 \cos (\omega t) ,
where \zeta and \omega _n are the damping ratio and undamped natural frequency of the system, respectively, while \omega is the forcing frequency. The amplitude of the forced steady state response of this system is given by [(1-r^2)^2+(2\varsigma r)^2]^{-\frac{1}{2}} , where r=\omega /\omega _n . The peak amplitude of this response occurs at a frequency \omega =\omega _p . If \omega _d denotes the damped natural frequency of this system, which one of the following options is true?
A
\omega _p \lt \omega _d \lt \omega _n
B
\omega _p = \omega _d \lt \omega _n
C
\omega _d \lt \omega _n = \omega _p
D
\omega _d \lt \omega _n \lt \omega _p
GATE ME 2022 SET-1      Vibration
Question 7 Explanation: 
The relation between the frequency at peak and natural frequency is given
\omega _p=\omega _n\sqrt{1-2\zeta ^2}
\omega _p \gt \omega _n
The relation between the damped frequency and natural frequency
\omega _d=\omega _n\sqrt{1-\zeta ^2}
\omega _d \gt \omega _n
\frac{\omega _p}{\omega _d}=\frac{\sqrt{1-2\zeta ^2}}{\sqrt{1-\zeta ^2}} \lt 1
Therefore, \omega _p \lt \omega _d \lt \omega _n
Question 8
A planar four-bar linkage mechanism with 3 revolute kinematic pairs and 1 prismatic kinematic pair is shown in the figure, where AB\perp CE and FD\perp CE. The T-shaped link CDEF is constructed such that the slider B can cross the point D, and CE is sufficiently long. For the given lengths as shown, the mechanism is

A
a Grashof chain with links AG, AB, and CDEF completely rotatable about the ground link FG
B
a non-Grashof chain with all oscillating links
C
a Grashof chain with AB completely rotatable about the ground link FG, and oscillatory links AG and CDEF
D
on the border of Grashof and non-Grashof chains with uncertain configuration(s)
GATE ME 2022 SET-1      Planar Mechanisms
Question 8 Explanation: 
The given mechanism is

As we know sliding pair is a special. Case of turning pair with infinite lengths limit. So the equivalent diagram would be. Since two parallel lines meets at infinite point O, 2O_2 are same.

l_1=3 cm shortest line,
l_2=5cm
l_3=l_{3(\infty )}=L_x+3 , longest link
l_4=l_{4(\infty )}=L_x+1.5
For Grashoff's rule to satisfy
l_1+l_3 \leq l_2 +l_4
3+L_x+3 \leq 5+L_x+1.5
6 \leq 6.5
LHS is less than RHS.
Hence, Grashoff's rule is satisfied in this mechanism. Since shortest link is fixed. It will be a double crank mechanism.
Question 9
A rigid uniform annular disc is pivoted on a knife edge A in a uniform gravitational field as shown, such that it can execute small amplitude simple harmonic motion in the plane of the figure without slip at the pivot point. The inner radius r and outer radius R are such that r^2=R^2/2, and the acceleration due to gravity is g. If the time period of small amplitude simple harmonic motion is given by T=\beta \pi \sqrt{R/g}, where \pi is the ratio of circumference to diameter of a circle, then \beta=________ (round off to 2 decimal places).

A
5.25
B
2.65
C
3.75
D
9.86
GATE ME 2022 SET-1      Vibration
Question 9 Explanation: 


Considering the differential ring mass of differential ring
\begin{aligned} d_m&=\frac{\text{mass of ring}}{\text{volume of ring}}(dV)\\ &=\frac{M}{\pi(R^2-r^2)t}2 \pi ada \times t\\ &=\frac{2Mada}{(R^2-r^2)}\\ I_0&=\bar{I}=\int a^2 d_m\\ &=\int_{r}^{R}\frac{2Ma^3da}{(R^2-r^2)}\\ &=\frac{2M}{(R^2-r^2)} \times \left [ \frac{a^4}{4} \right ]_r^R\\ &=\frac{M}{2}\frac{R^4-r^4}{R^2-r^2}\\ &=\frac{M}{2}{R^2+r^2}\\ I&=\bar{I}+mr^2\\ &=\frac{M}{2}(R^2+r^2)+Mr^2=\frac{5MR^2}{4} \end{aligned}


Taking moments about A
\begin{aligned} \Sigma M_A&=0\\ I\ddot{\theta }+Mg(r \sin \theta )&=0\\ \frac{5MR^2}{4}\ddot{\theta }+Mg\left ( \frac{R}{\sqrt{2}} \right )\theta &=0\;\;(\because \sin \theta \cong \theta )\\ \ddot{\theta }+\frac{4g}{5\sqrt{2}R}\theta &=0\\ \omega _n&=\sqrt{\frac{4g}{5\sqrt{2}R}}\\ T&=\frac{2 \pi}{\omega _n}= \pi\left ( 2 \times \sqrt{\frac{5\sqrt{2}}{5} \times \frac{R}{g}} \right )\\ \beta &=2.65 \end{aligned}
Question 10
The figure shows a schematic of a simple Watt governor mechanism with the spindle O_1O_2 rotating at an angular velocity \omega about a vertical axis. The balls at P and S have equal mass. Assume that there is no friction anywhere and all other components are massless and rigid. The vertical distance between the horizontal plane of rotation of the balls and the pivot O_1 is denoted by h . The value of h=400 mm at a certain \omega . If \omega is doubled, the value of h will be _________ mm.

A
50
B
100
C
150
D
200
GATE ME 2022 SET-1      Gyroscope
Question 10 Explanation: 
h_1 = 400 mm, h_2 = ?
\omega _1=\omega \;\;\;\omega _2=2\omega
For Watt governor,
\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}


There are 10 questions to complete.

9 thoughts on “Theory of Machine”

    • 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.

      Thank you for your support.

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