Question

In a certain slider-crank mechanism, lengths of crank and connecting rod are equal. If the crank rotates with a uniform angular speed of 14 rad/s and the crank length is 300 mm, the maximum acceleration of the slider (in m/s²) is ___________

Answer 1

Concept:

In a slider crank-mechanism, 

The piston displacement: \({x_p} = r\left( {1 - \cos θ } \right)\)

Piston velocity: \({v_p} = \omega r\left( {\sin θ + \frac{{\sin 2θ }}{{2n}}} \right)\)

Piston acceleration: \({a_p} = {\omega ^2}r\left( {\cos θ + \frac{{\cos 2θ }}{n}} \right)\),  where n is equal to l/r 

l length of connecting rod in m and r is crank radius in m.

for maximum acceleration, θ = 0^° \({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{n}} \right)\)

Calculation:

Given, l = r = 300 mm = 0.3 m, ω = 14 rad/s

\({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{n}} \right)\)

n = 1, as l = r = 0.3 m

thus, \({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{1}} \right) = 2{\omega ^2}r\)

\({a_{max}} = 2 \times {14^2} \times 0.3 = 117.6\;m/{s^2}\)