Question
Download Solution PDFThe given figure shows a quick return mechanism. The crank OA rotates clockwise uniformly. OA = 2 cm, OO' = 4 cm. The ratio of time taken for forwarding motion to that for return motion is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
A cutting stroke occurs when the crank rotates through an angle β and the return stroke occurs when the crank rotates through an angle α or (360° - β) in the clockwise direction.
\(\frac{{Time~of~cutting~stroke}}{{Time~of~return~stroke}} = \frac{β }{α } = \frac{β }{{\left( {360^\circ - β } \right)}}\)
Calculation:
Given:
Length of the crank, AO = 2 cm
Distance between the center of the crank and slotted lever, OO' = 4 cm
Consider ΔOO'A,
\(cos \frac{α}{2}=\frac{OA}{OO'}\)
\(cos\frac{α}{2}=\frac{2}{4}=\frac{1}{2}\)
Therefore, α = 120°
The cutting stroke angle β,
β = 360 - α
β = 360 - 120 = 240°
\(\frac{{Time~of~cutting~stroke}}{{Time~of~return~stroke}} = \frac{β }{α } = \frac{240}{120}=2\)
Last updated on May 28, 2025
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