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In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to

In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to Correct Answer minimize the effect of secondary forces

As l increases FS decreases Hence the ratio is kept large to avoid secondary forces for each balancing.

Related Questions

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 the equivalent length L of a simple pendulum swung about an axis.
For a steam engine, the following data is given: 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 calculate inertia force at θ=30 degrees from IDC.
From the data given: 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 correcting couple in N-m?
In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine linear velocity of the midpoint of the connecting rod in m/s. Crank angle = 45° from IDC.
In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine acceleration of the midpoint of the connecting rod in m/s2. Crank angle = 45° from IDC.
In a slider crank mechanism, the length of the crank and connecting rod are 180 mm and 540 mm respectively. The crank position is 45° from inner dead centre. The crank shaft speed is 450 r.p.m. (clockwise), calculate angular velocity of the connecting rod in rad/s.
A hollow rod has frictionless inner walls. Inside the rod are two small spheres that are kept on either side of the centre of rod. The rod is rotated about its central axis which is perpendicular to the plane of rotation. If the rod had been rotated by an impulsive torque which gave it an instantaneous angular velocity of 5rad/s, what will be the angular velocity after some finite but long time ‘t’? Assume no external forces act on rod after the impulse. Also state what will happen to the spheres in the centre of the rod. The rod has a mass =2kg & length =10cm. The two spheres have a mass =1kg each.
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)