Question

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of 1.25 rad/s. The opening between a successive blades is equal to the width of a blade. A golf ball (diameter 4.50 x 10^-2 m) is just passing by one of the rotating blades. What must be the minimum speed of the ball so that it will not be hit by the next blade?

Answer 1

To understand the passage between two blades, it is required

to travel the distance of the circumference equivalent to the

segment of the diameter that exists between them,

$v = \frac{d_{ball}}{\Delta t}$

Where

$d_{ball} =$Ball diameter

$\Delta t=$ Space time

So the angle swept out by either a blade or a space is:

$\theta = 2\pi / 16 = \pi / 8 rad.$

Through the angular velocity

$\omega = \frac{\theta} {t}$

$t = \frac{\theta}{\omega}$

$t= \frac{\pi /8}{1.25} = 0.3141s$

So,

$v = 4.50*10^-2m / 0.3141 s = 0.1432m/s$