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.

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. Correct Answer 17.824

α = -ω2.sinθ/n = 13.12.sin3θ/5 = 17.161 rad/s2.

Related Questions

The primary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by (where m = Mass of reciprocating parts, $$\omega $$ = Angular speed of crank, r = Radius of crank, $$\theta $$ = Angle of inclination of crank with the line of stroke and n = Ratio of the length of connecting rod to radius of crank)
The secondary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by (where m = Mass of reciprocating parts, $$\omega $$ = Angular speed of crank, r = Radius of crank, $$\theta $$ = Angle of inclination of crank with the line of stroke and n = Ratio of the length of connecting rod to radius of crank)