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

Force which does not act on the connecting rod is ______

a) Weight of connecting rod

b) Inertia force of connecting rod

c) Radial force

d) Coriolis force

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Answer: d

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

Inertia forces on the reciprocating parts acts along the line of stroke.

a) True

b) False

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Answer: a

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.

When mass of the reciprocating parts is neglected then the inertia force is _____

a) Maximum

b) Minimum

c) 0

d) Not defined

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Answer: c

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.

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

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Answer: b

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

ω=13.1 rad/s r=0.3m

substituting these values will give Fi = 19064 N.

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

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Answer: b

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.

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

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Answer: c

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.

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

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Answer: b

Explanation: α = -ω2.sinθ/n

= 13.12.sin3θ/5

= 17.161 rad/s2.

Torque due to weight of the connecting rod affects the torque due to connecting rod.

a) True

b) False

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Answer: b

Explanation: The torque due to connecting rod remains same irrespective of the torque caused by the weight of the connecting rod.