Question

A quick-return mechanism is to be designed, where the outward stroke must consume 1.2 s and the return stroke 0.8 s. If the cycle time is 2.0 s/rev, what is the speed at which the mechanism should be driven ?
1. 10 rev/s
2. 30 rev/s
3. 10 rev/min
4. 30 rev/min
5.

Answer 1

Correct Answer - Option 4 : 30 rev/min

Concept:

Quick Return Ratio (QRR):

\(\begin{array}{l} QRR = \frac{{Time\;for\;forward\;stroke}}{{Time\;for\;return\;stroke}} = \frac{{{T_f}}}{{{T_r}}} \end{array}\)

where T = cycle-time = T_f + T_r

The relation between time-period and crank speed in rpm is given by:

\(\omega=\frac{2\pi}{T}\)

Calculation:

Given:

T_f = 1.2 sec, T_r = 0.8 sec, T = T_f + T_r = 2 sec/rev

We know that

\(\omega =\frac{2\pi N}{60}\)

Therefore, 

\(\omega=\frac{2\pi}{T}\Rightarrow \frac{2\pi N}{60}=\frac{2\pi}{T}\)

\(\frac{N}{60}=\frac{1}{T}\)

\({N}{}=\frac{60}{T}\Rightarrow \frac{60}{2}=30\;rev/min\)