An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin. What will be the new radius of gyration?

Question

An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin. What will be the new radius of gyration?

asked Sep 22, 2022 in Machine Dynamics by Grayson (34.3k points)

An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin. What will be the new radius of gyration?

(a) 0.212m

(b) 0.122m

(c) 0.1m

(d) 0.145m

I got this question by my school teacher while I was bunking the class.

Question is from Correction Couple to be Applied to Make the Two Mass Systems Dynamically Equivalent in section Inertia Forces in Reciprocating Parts of Machine Dynamics

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Fred · Answer 1 · 2022-10-14T07:32:35+0000

Correct answer is (b) 0.122m

Easiest explanation: Let l3 be the distance between the masses, then l3 = l- l1 = 15cm

If K1 is the new radius of gyration then, K1^2= l1. l3 = 0.1×0.15 = 0.015 m^2

Therefore, K1 = 0.122 m.

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