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, if the angular acceleration of the rod is 23 000 rad/s^2, then find the correction couple in N-m.

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, if the angular acceleration of the rod is 23 000 rad/s^2, then find the correction couple in N-m.

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, if the angular acceleration of the rod is 23 000 rad/s^2, then find the correction couple in N-m.

(a) 140.2

(b) 133.4

(c) 136.8

(d) 135.6

The question was posed to me during an interview.

Question is taken 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 · 2022-10-14T04:18:01+0000

Correct choice is (b) 133.4

To explain: We know that correction couple is given by T’ = m.(k1^2 – kg^2).α

T’ = 2. (0.015 – 0.11^2). 23000

= 133.4 N-m.

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