# Planar Mechanisms

 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 $\omega _2 =5$rad/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   Theory of Machine
Question 1 Explanation:

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 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   Theory of Machine
Question 2 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 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 Explanation:

\begin{aligned} F &=3(6-1)-2 \times 7-0 \\ &=15-14=1 \end{aligned}
 Question 8
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 8 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 9
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 9 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 10
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 10 Explanation:
$S+L \lt P+Q$
In this case, shortest link is fixed and hence, the
resulting mechanism is double crank mechanism.
There are 10 questions to complete.

### 5 thoughts on “Planar Mechanisms”

1. I think questions 1 diagram are not matchable

• Thank you NAVEENKUMAR,
We have updated the figure.

2. 10 the questions not understand plz anyone explain

• Sorry 20th

3. In question no 36, I think the answer should be B, bcoz cam and follower is the mechanism with DOF 1 having 3 links with 2 lower pairs & 1 higher pair.
if in the problem they ask DOF 1 with lower pair only then minimum 4 links are required (from grablur’s eqn)