Don't know where to start!

Hey, any ideas on this double starred, hard question :(

A golf ball passing trhough a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of 1.25 rad/s clockwise. The opening between successive blades is equal to the width of a blade. A golf ball of diameter 4.50x10^-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?

Ok, my thought is, I need to find time for a blade to rotate theta, and I need the arc length for that. With arc length I have some way to compare the diameter and the space opening between the ball and blade. Lost from that point... :yuck:

Any ideas, Great!

NateTG
Homework Helper
How far does the ball have to travel?
How big (as an angle) are the gaps in the windmill?

Ok, knowing w, I think I can use theta= w(avg) x time . I don't have time. I have 1.25rad/s. Hold it is this approach right. Ok I have 1.25 rad, 2pie is a full revolution or 360 degrees. There is a 5.02 factor between 1.25 rad and 2 pie. Ok, 1.25 rad is equal to 71.61 degrees. I multiply that by 5.026 and divide by 16 (total equal proportions including space and blade) and I get roughly 22.5 degrees for each gap?

As for how far does the ball have to travel, is that relevant? I mean it just has to travel pass the width of the blade.

NateTG
Homework Helper
You should be looking at things like: There are 8 blades, so there are 8 gaps. That means that each gap is going to be 1/16 of a revolution, or an angle of $$\frac{2\pi}{16}=\frac{\pi}{8}$$.

Then you can determine the time that the windmill takes to turn that far and so on.

ok, so my way was also corrrect than ... 22.5 degrees is equal 0.39 rad which is equal to pie/8. Now for the next part, time. I'm thinking we can use theta= 1/2 (w avg )t. Since the mill will spin always at 1.25 rad/s the w avg is still 1.25rad/s. Theta = 0.39 rad, I substitute and solve for t.

t = 0.628 s.
Thinking...

ok I figure: need the angular velocity of ball (w) then we could use this value and substitute it into V=rw? so diameter we know is 0.0450, d/2 = r. So, w = theta/time. So, 0.39/0.628 gives 0.621 rad/s. Substitute it into V=rw, V=0.01397m/s?

NateTG
Homework Helper
$$\theta=\omega_{avg}t$$
(There's no $$\frac{1}{2}$$ you may be thinking of $$\theta=\alpha_{avg}t$$.)

Then the ball needs to travel it's diameter in the time. so $$v=\frac{d}{t}$$

(This would be a lot easier to explain with drawings.)

Everything sounds good, except I'm not sure I understand the concept, the ball needs to travel it's diameter?

NateTG
Homework Helper
I can try to do ascii art:
Imagine you're looking at the windmill from the top
Then the beginning of the ball crossing will look like this
Code:
 |  Blade (moving down on screen)

|
| Blade (moving down)

And the end of the ball crossing will look like
Code:
 |
|o--> (Ball having crossed)

|
So, if the ball barely scrapes the tail end of the blade before it starts crossing, and barely gets scraped by the blade at the end of the crossing, then it's traveled it's diameter in distance.

Well, the diagrams make some sense, excpet I don't see how both parts of the mill can move down, shouldn't it be rotating so that one side goes up while the other comes down. But yes, I get it now, I thought it was the travel distance before and reaches the mill plus the distance to get pass it.

Thanks!

NateTG