Theory of Machine


Question 1
The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity \omega _1. A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor.
What is the angular velocity \omega _1 of the bar PQ?

A
\omega _2=2\omega _1
B
\omega _2=\omega _1
C
\omega _2=0.5\omega _1
D
\omega _2=0.25\omega _1
GATE ME 2023      Displacement, Velocity and Acceleration
Question 1 Explanation: 


As Disc (Link 2) roll without slipping on Link 1 (surface) so their I-centre of rotation lies on point of contact to avoid relative motion in between them. So \mathrm{I}_{12} lies on point of contact.
Link 2 and 3 have instantaneous center of rotation at \mathrm{I}_{13} due to fixed surface of Link 1.

As per point Plies on link 2 and 3 both so, point P has same velocity for Link 2 and 3 both.
As per Kennedy's theorem, if three plane bodies have relative motion among themselves, their I-centre must lie on a straight line. So, \mathrm{I}_{23} must be on line joining \mathrm{I}_{13} and \mathrm{I}_{12} at point P.
By similar \triangle \mathrm{PI}_{12} \mathrm{~B} and \Delta \mathrm{Pl}_{13} \mathrm{~A}, we have
\frac{\mathrm{l}_{12} \mathrm{P}}{\mathrm{l}_{12} \mathrm{~B}}=\frac{\mathrm{Pl_{13 }}}{\mathrm{PA}}
or, \quad \frac{\mathrm{I}_{12} \mathrm{I}_{23}}{2}=\frac{\mathrm{I}_{23} \mathrm{l}_{13}}{8}
or, \mathrm{I}_{12} \mathrm{l}_{13}=0.25 \mathrm{I}_{23} \mathrm{l}_{13} \;\;\;...(i)
Velocity of point P is same w.r.t. all instantaneous rotation so,
\omega_{1} \times I_{12} I_{23}=\omega_{2} \times I_{23} l_{13}
By equation (i), we get
\omega_{2}=0.25 \omega_{1}
Question 2
The figure shows a block of mass m = 20 kg attached to a pair of identical linear springs, each having a spring constant k = 1000 N/m. The block oscillates on a frictionless horizontal surface. Assuming free vibrations, the time taken by the block to complete ten oscillations is _________ seconds. (Rounded off to two decimal places)
Take \pi= 3.14

A
2.36
B
5.35
C
8.25
D
6.28
GATE ME 2023      Vibration
Question 2 Explanation: 
K_{eq}=K+K=2K=2 \times 1000=2000 \mathrm{~N} / \mathrm{m}
\mathrm{m}=20 \mathrm{~kg}
So, \quad \omega_{n}=\sqrt{\frac{K_{\text {eq }}}{m}}=\sqrt{\frac{2000}{20}}=10 \mathrm{rad} / \mathrm{sec}
So, \quad T_{n}=\frac{2 \pi}{\omega_{n}}=\frac{2 \pi}{10}
Time taken for 10 oscillations =10 \times T_{n}=10 \times \frac{2 \pi}{\omega_{n}}=10 \times \frac{2 \pi}{10}=6.28 \mathrm{sec}


Question 3
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)

A
52
B
48
C
96
D
112
GATE ME 2023      Gear and Gear Train
Question 3 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 4
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 4 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 5
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 5 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





There are 5 questions to complete.

12 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.

      Reply

  7. In Theory Of Machine Q. No. 34. Explanation is given wrong and also answer given in the explanation do not match with the correct answer given in the option. Kindly correct it.

    Reply

Leave a Comment