- #1
a.mlw.walker
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OK. I have been fighting with my pen and paper for hours trying to fathom this:
I have been thinking about that childs game where there is a post, and two players. On the post is a string with a tennis ball attached at the top, each player has a tennis raquet and they hit the ball around the ball to each other.
As they hit the ball harder the ball rises, and therefore its radius increases. So the
-centepetal force is increasing (or centrefugal but we know that's fiction...) and therefore the force is more and more overcoming the mg of the ball, hence it rises towards horizontal.
If there was only one player, and he hits the ball around and around, until it is quite close to the horizontal (it will never be completely horizontal), and you wanted to work out where the ball will be in 1.5 seconds, what do you have to take into account.
Is the acc (change in speed/change in time)?
deceleration is constant and therefore we can work its distance traveled in 1.5 seconds from constant acc equations?
and r is a function of what?
Its a simple game but so hard to calculate because r changes as ball decelerates.
I am right in thinking though that as long as centrepetal force is greater than mg the ball will stay airborne?
I bet there are others out there who notice my little predicament...
can anyone give me a mathematical explanation for what's going on, I understand in words but not equations...
There is also a constant speed at which the tennis ball will no longer spin, but fall, when its centrepetal force goes less than mg, which i would also like to calculate...
Thanks
Alex
I have been thinking about that childs game where there is a post, and two players. On the post is a string with a tennis ball attached at the top, each player has a tennis raquet and they hit the ball around the ball to each other.
As they hit the ball harder the ball rises, and therefore its radius increases. So the
-centepetal force is increasing (or centrefugal but we know that's fiction...) and therefore the force is more and more overcoming the mg of the ball, hence it rises towards horizontal.
If there was only one player, and he hits the ball around and around, until it is quite close to the horizontal (it will never be completely horizontal), and you wanted to work out where the ball will be in 1.5 seconds, what do you have to take into account.
Is the acc (change in speed/change in time)?
deceleration is constant and therefore we can work its distance traveled in 1.5 seconds from constant acc equations?
and r is a function of what?
Its a simple game but so hard to calculate because r changes as ball decelerates.
I am right in thinking though that as long as centrepetal force is greater than mg the ball will stay airborne?
I bet there are others out there who notice my little predicament...
can anyone give me a mathematical explanation for what's going on, I understand in words but not equations...
There is also a constant speed at which the tennis ball will no longer spin, but fall, when its centrepetal force goes less than mg, which i would also like to calculate...
Thanks
Alex
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