Planar Mechanisms

Question 1
A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are V_P \; and \; V_R, in the x and y directions, respectively. The magnitude of the angular velocity of the body is
A
2V_R
B
2V_P
C
V_R/\sqrt{3}
D
V_P/\sqrt{3}
GATE ME 2019 SET-2   Theory of Machine
Question 1 Explanation: 
\begin{array}{l} \Rightarrow \mathrm{V}_{\mathrm{R}}=(\mathrm{IR}) \omega \\ \Rightarrow \omega=\frac{\mathrm{V}_{\mathrm{R}}}{(\mathrm{IR})} \\ \Rightarrow \omega \times \frac{\mathrm{V}_{\mathrm{R}}}{\frac{1}{2}} \\ \Rightarrow \omega=2 \mathrm{V}_{\mathrm{R}} \end{array}

Question 2
In a four bar planar mechanism shown in the figure, AB = 5 cm, AD = 4 cm and DC = 2 cm. In the configuration shown, both AB and DC are perpendicular to AD. The bar AB rotates with an angular velocity of 10 rad/s. The magnitude of angular velocity (in rad/s) of bar DC at this instant is
A
0
B
10
C
15
D
25
GATE ME 2019 SET-1   Theory of Machine
Question 2 Explanation: 
\begin{array}{l} \mathrm{AB}=5 \mathrm{cm} \\ \mathrm{AD}=4 \mathrm{cm} \\ \mathrm{DC}=2 \mathrm{cm} \\ \omega_{\mathrm{AB}}=10 \mathrm{rad} / \mathrm{s} \\ \because A \mathrm{B} \| \mathrm{DC} \\ \therefore A B \cdot \omega_{\mathrm{AB}}=\mathrm{DC} \cdot \omega_{\mathrm{DC}} \\ 5 \times 10=2 \times \omega_{\mathrm{DC}} \\ \omega_{\mathrm{DC}}=25 \mathrm{rad} / \mathrm{s} \end{array}
Question 3
In a slider-crank mechanism, the lengths of the crank and the connecting rod are 100mm and 160mm, respectively. The crank is rotating with an angular of 10 radian/s counter-clockwise. The magnitude of linear velocity (in m/s) of the piston at the instant corresponding to the configuration shown in the figure is_____.
A
2
B
1.5
C
1
D
0.5
GATE ME 2017 SET-2   Theory of Machine
Question 3 Explanation: 


After plotting I-centres

Here, I_{23} and I_{24} will come at same point
Applying angular Velocity Theorem at I_{24}
\begin{aligned} \therefore \quad \omega_{2}\left(I_{24} I_{12}\right) &=V_{4}=V_{B} \\ V_{B} &=\omega_{2}\left(I_{24} I_{12}\right)=10 \times 0.1 \\ \left(I_{24} I_{12}=100 \mathrm{mm}\right.&=0.1 \mathrm{m}=A B) \\ V_{B} &=1 \mathrm{m} / \mathrm{s} \end{aligned}
Question 4
For an inline slider-crank mechanism, the lengths of the crank and connecting rod are 3m and 4m, respectively. At the instant when the connecting rod is perpendicular to the crank, if the velocity of the slider is 1m/s, the magnitude of angular velocity (upto 3 decimal points accuracy) of the crank is _________ radian/s.
A
0.222
B
0.266
C
0.298
D
0.316
GATE ME 2017 SET-1   Theory of Machine
Question 4 Explanation: 


\begin{aligned} \text { Here } & V_{B}=1 \mathrm{m} / \mathrm{s} \\ & \omega_{\text {crank }}=? \end{aligned}
Let us draw the I-centres


Applying angular velocity theorem at I_{24}
\begin{array}{l} \therefore \omega_{2}\left(I_{24} I_{12}\right)=V_{4}=V_{B}=1 \\ \omega_{2}\left(I_{24} I_{12}\right)=1 \end{array}


\begin{aligned} \theta &=\tan ^{-1}(4 / 3)=53.13010 \\ \alpha &=90^{\circ}-\theta=36.869 \\ \cos \alpha &=\frac{3}{I_{12} I_{24}} \\ I_{12} I_{24} &=\frac{3}{\cos 36.869}=3.7499 \mathrm{meter} \\ \text{Now}, & \omega_{2}(3.7499)=1 \\ \omega_{2} &=\frac{1}{3.7499}=0.266 \mathrm{rad} / \mathrm{s} \end{aligned}
Question 5
The number of degrees of freedom of the linkage shown in the figure is
A
-3
B
0
C
1
D
2
GATE ME 2015 SET-3   Theory of Machine
Question 5 Explanation: 


\begin{aligned} F &=3(6-1)-2 \times 7-0 \\ &=15-14=1 \end{aligned}
Question 6
Consider a slider crank mechanism with nonzero masses and inertia. A constant torque \tau is applied on the crank as shown in the figure. Which of the following plots best resembles variation of crank angle, \theta versus time
A
A
B
B
C
C
D
D
GATE ME 2015 SET-1   Theory of Machine
Question 6 Explanation: 
\begin{aligned} \tau(\text { torque }) &=I \ddot{\theta} \\ \tau &=\text { constant } \\ I &=\text { constant } \\ \tau &=I \frac{d^{2} \theta}{d t^{2}} \\ \frac{d^{2} \theta}{d t^{2}} &=\frac{\tau}{I} \\ \frac{d \theta}{d t} &=\frac{\tau}{I} t+C_{1} \\ \theta &=\frac{\tau}{I} \frac{t^{2}}{2}+C_{1} t+C_{2}\\ \theta=f(t)&= \text{ parabolic function}\\ \end{aligned}
But practically graph will be fluctuating about ideal parabolic curve.
Question 7
In a statically determinate plane truss, the number of joints (j) and the number of members (m) are related by
A
j=2m-3
B
m=2j+1
C
m=2j-3
D
m=2j-1
GATE ME 2014 SET-4   Theory of Machine
Question 7 Explanation: 
A simple truss is formed by enlarging the basic truss element which contains three members and three joints, by adding two additional members for each additional joint, so the total number of members m in a simple truss is given by
m=3+2(j-3)=2 j-3
where m= number of members
j= total number of joints (including those attached to the supports)
Question 8
A 4-bar mechanism with all revolute pairs has link lengths l_f=20mm, l_{in}=40mm, l_{co}=50mm and l_{out}=60mm. The suffixes 'f', 'in', 'co' and 'out' denote the fixed link, the input link, the coupler and output link respectively. Which one of the following statements is true about the input and output links?
A
Both links can execute full circular motion
B
Both links cannot execute full circular motion
C
Only the output link cannot execute full circular motion
D
Only the input link cannot execute full circular motion
GATE ME 2014 SET-2   Theory of Machine
Question 8 Explanation: 
S+L \lt P+Q
In this case, shortest link is fixed and hence, the
resulting mechanism is double crank mechanism.
Question 9
A rigid link PQ is 2 m long and oriented at 20^{\circ} to the horizontal as shown in the figure. The magnitude and direction of velocity V_{Q} , and the direction of velocity V_{P} are given. The magnitude of V_{P}(in m/s) a t this instant is
A
2.14
B
1.89
C
1.21
D
0.96
GATE ME 2014 SET-1   Theory of Machine
Question 9 Explanation: 


\begin{aligned} \therefore \; \omega _{PQ}&=\frac{V_P}{PI_{PQ}}=\frac{V_Q}{QI_{PQ}} \\ PI_{PQ}&=\text{Distance between P and }I_{PQ} \\ &= 2 \cos 70^{\circ}+ 2 \sin 70^{\circ} \times \tan 45^{\circ}\\ QI_{PQ} &= \text{Distance between Q and }I_{PQ} \\ &=\sqrt{2}\times 2 \times \sin 70^{\circ} \\ &= 2.66m\\ \therefore \;\; \frac{V_P}{2.56} &=\frac{1}{2.66}\\ V_P&=\frac{2.56}{2.66}=0.96\; m/s \end{aligned}
Question 10
A planar closed kinematic chain is formed with rigid links PQ = 2.0 m, QR = 3.0 m, RS = 2.5 m and SP = 2.7 m with all revolute joints. The link to be fixed to obtain a double rocker (rockerrocker) mechanism is
A
PQ
B
QR
C
RS
D
SP
GATE ME 2013   Theory of Machine
Question 10 Explanation: 
\begin{array}{l} P Q=2.0 \mathrm{m}, Q R=3.0 \mathrm{m}, R S=2.5 \mathrm{m} \\ S P=2.7 \mathrm{m} \end{array}
(i) If shortest line PQ=2.0m is fixed, then we will give double crank.
(ii) If link adjacent to shortest is fixed, then we will get crank-rocker mechanism.
(iii) If we fix coupler 'RS', we will get double rocker mechanism.
There are 10 questions to complete.

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