Bissoy.com
Doctors Medicines MCQs Q&A

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.

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. Correct Answer 11.1

n = l/r : 540/180 = 3 angular velocity is given by ω.cos θ/n = (450*2pi / 60).cos45/3 = 11.1 rad/s.

Related Questions

In a slider crank mechanism, the length of the crank and connecting rod are 150 mm and 600 mm respectively. The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m. (clockwise). Velocity of the slider is ________
Consider the following statements: In a slider-crank mechanism, the slider is at its dead centre position when the (i) slider velocity is zero (ii) slider velocity is maximum (iii) slider acceleration is zero (iv) slider acceleration is maximum Which of the above statements are correct?
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.
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.
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.
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.
From the data given: The length of the crank and connecting rod are 150 mm and 600 mm The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m. Find the angular acceleration in rad/s2 of the connecting rod.
From the data given: The length of the crank and connecting rod are 150 mm and 600 mm The crank position is 60° from inner dead centre. The crank shaft speed is 400 r.p.m. Find the acceleration in m/s2 of the slider.
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.
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.