Discussion
\(180 - 2\sin^{-1}\left(\dfrac{r_1 - r_2}{x}\right)\)
The above expression belongs to
Answer: C
The expression \(180 - 2\sin^{-1}\frac{r_1 - r_2}{x}\) gives the angle of contact for an open belt drive. In an open belt drive the two pulleys of radii \(r_1\) and \(r_2\) are at a centre distance x, and the angle of contact on the smaller pulley is the angle subtended by the belt, which is \(180^{\circ} - 2\alpha\) where \(\sin\alpha = \frac{r_1 - r_2}{x}\). In a crossed belt drive the angle is \(180^{\circ} + 2\alpha\) with \(\sin\alpha = \frac{r_1 + r_2}{x}\), and the key gives the angle of contact rather than the angle of lap for the open belt.
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