A connecting rod of an internal combustion engine has a mass of 1.5 kg and the length of the rod is 250 mm. The centre of gravity of the rod is located at a distance of 100 mm from the gudgeon pin. The radius of gyration about an axis through the centre of gravity perpendicular to the plane of

Question

A connecting rod of an internal combustion engine has a mass of 1.5 kg and the length of the rod is 250 mm. The centre of gravity of the rod is located at a distance of 100 mm from the gudgeon pin. The radius of gyration about an axis through the centre of gravity perpendicular to the plane of rotation is 110 mm. Find the equivalent dynamical system if only one of the masses is located at the gudgeon pin.

If the connecting rod is replaced by two masses, one at the gudgeon pin and the other at the crank pin and the angular acceleration of the rod is 24000 rad/s² clockwise, determine the correction couple applied to the system to reduce it to a dynamically equivalent system.

Accepted Answer

[latex] ext { Here } m=1.5 kg , L=250 mm , a=100 mm , K=110 mm [/latex].

Now [latex] a b=K^{2} [/latex].

[latex] b=\frac{110^{2}}{100}=121 mm [/latex].

[latex] ext { Let } m= ext { mass at the gudgeon pin } [/latex].

[latex] m_{b}= ext { mass at the crank pin } [/latex].

Then [latex] m_{a}=\frac{m_{b}}{L}=\frac{1.5 imes 121}{250}=0.72 kg [/latex].

[latex] m_{b}=1.5-0.72=0.78 kg [/latex].

Correction couple
Now [latex] a=100 mm , c=150 mm [/latex].

[latex] K_{1}^{2}=a c=100 imes 150=15000 [/latex].

[latex] K_{1}=122.47 mm [/latex].

Correction couple, [latex] T_{o}=m\left(K_{1}^{2}-K^{2} ight) \alpha [/latex].

[latex] =1.5(15000-12100) imes 24000 imes 10^{-6} [/latex].

= 104.4 N m.

Question 10.44: A connecting rod of an internal combustion engine has a mass...

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