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).

1.25 | |

0.75 | |

2.25 | |

3.75 |

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 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 Grashof chain with links AG, AB, and CDEF completely rotatable about the ground link FG | |

a non-Grashof chain with all oscillating links | |

a Grashof chain with AB completely rotatable about the ground link FG, and oscillatory links AG and CDEF | |

on the border of Grashof and non-Grashof chains with uncertain configuration(s) |

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.

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

2V_R | |

2V_P | |

V_R/\sqrt{3} | |

V_P/\sqrt{3} |

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

0 | |

10 | |

15 | |

25 |

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_____.

2 | |

1.5 | |

1 | |

0.5 |

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.

0.222 | |

0.266 | |

0.298 | |

0.316 |

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

-3 | |

0 | |

1 | |

2 |

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 | |

B | |

C | |

D |

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.

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

j=2m-3 | |

m=2j+1 | |

m=2j-3 | |

m=2j-1 |

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)

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?

Both links can execute full circular motion | |

Both links cannot execute full circular motion | |

Only the output link cannot execute full circular motion | |

Only the input link cannot execute full circular motion |

Question 10 Explanation:

S+L \lt P+Q

In this case, shortest link is fixed and hence, the

resulting mechanism is double crank mechanism.

In this case, shortest link is fixed and hence, the

resulting mechanism is double crank mechanism.

There are 10 questions to complete.

I think questions 1 diagram are not matchable

Thank you NAVEENKUMAR,

We have updated the figure.

10 the questions not understand plz anyone explain

Sorry 20th

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)