Question

An open V-belt is wrapped around V-pulleys having effective diameters of 0.25 m and 0.65 m, and their centres are 1 m apart. Assuming ideal conditions, the wrap angle in degrees for the smaller pulley is _____. \textit{[Round off to two decimal places.]}

Hint:

The wrap angle for a V-belt is influenced by the distance between pulleys and their diameters. Use the formula involving the sine function for accurate calculation.

Answer 1

Correct Answer: 156

The wrap angle for the smaller pulley in a V-belt drive can be calculated using the following formula: \[ \theta = 2 \times \arcsin \left( \frac{d_2 - d_1}{2L} \right) \] where:
- \( d_1 = 0.25 \, \text{m} \) is the diameter of the smaller pulley,
- \( d_2 = 0.65 \, \text{m} \) is the diameter of the larger pulley,
- \( L = 1 \, \text{m} \) is the distance between the centres of the pulleys.
Substitute the values into the formula: \[ \theta = 2 \times \arcsin \left( \frac{0.65 - 0.25}{2 \times 1} \right) = 2 \times \arcsin \left( \frac{0.4}{2} \right) = 2 \times \arcsin(0.2). \] Using a calculator: \[ \arcsin(0.2) \approx 11.54^\circ. \] Thus, \[ \theta = 2 \times 11.54^\circ = 23.08^\circ. \] The wrap angle for the smaller pulley is approximately \( \boxed{23.08^\circ} \) (rounded to two decimal places).