+1 vote
in Machine Dynamics by (34.3k points)
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

I have been asked this question by my school teacher while I was bunking the class.

Query is from Inertia Forces in a Reciprocating Engine Considering the Weight of Connecting Rod topic in section Inertia Forces in Reciprocating Parts of Machine Dynamics

1 Answer

0 votes
by (38.6k points)
Right choice is (b) 1.42 m

The best explanation: We know that equivalent length L is given by the expression

L = (Kg^2 + l1^2)/l1

Kg = 0.65m      l1 = 1m

therefore L = 1.42 m.

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