Engineering Questions with Answers - Multiple Choice Questions

# Machine Dynamics MCQ – Inertia Forces in a Reciprocating Engine Considering the Weight of Connecting Rod

1 - Question

Force which does not act on the connecting rod is ______
a) Weight of connecting rod
b) Inertia force of connecting rod
d) Coriolis force

Explanation: Since there is no accelerated frame of reference having different velocities, therefore coriolis force does not act on the connecting rod.

2 - Question

Inertia forces on the reciprocating parts acts along the line of stroke.
a) True
b) False

Explanation: The reciprocating parts of the engine experience an Inertia force of magnitude m.ω2.r(cosθ + cos2θ/n), this inertia force acts along the line of stroke.

3 - Question

When mass of the reciprocating parts is neglected then the inertia force is _____
a) Maximum
b) Minimum
c) 0
d) Not defined

Explanation: Inertia force for neglected mass of reciprocating parts is 0 as it depends on the mass, since it has a value, it is defined.

4 - Question

For a steam engine, the following data is given:
Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
calculate inertia force at θ=30 degrees from IDC.
a) 19000 N
b) 19064 N
c) 19032 N
d) 20064 N

Explanation: l1 = l – G.C = 1.5 – 0.5 = 1m
Fi = (Mr + G.C,Mc/l).ω2.r(cosθ + cos2θ/n)
Mr=300 Kg
Mc=250 kg
substituting these values will give Fi = 19064 N.

5 - Question

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find the equivalent length L of a simple pendulum swung about an axis.
a) 1.35 m
b) 1.42 m
c) 1.48 m
d) 1.50 m

Explanation: We know that equivalent length L is given by the expression
L = (Kg2 + l12)/l1
Kg = 0.65m l1 = 1m
therefore L = 1.42 m.

6 - Question

From the data given:
Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find the correcting couple in N-m?
a) 52.7
b) 49.5
c) 59.5
d)56.5

Explanation: The correction couple depends on equivalent length, l1 mass of connecting rod and angular position and velocity
Tc = Mc.l1.(l-L).(ω2.sin2θ/2n2)
substituting the values into the equation results in
Tc = 59.5 N-m.

7 - Question

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find angular acceleration of connecting rod in rad/s2.
a) 16.782
b) 17.824
c) 15.142
d) 17.161

Explanation: α = -ω2.sinθ/n
= 13.12.sin3θ/5

8 - Question

Torque due to weight of the connecting rod affects the torque due to connecting rod.
a) True
b) False