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}

There are 5 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)