Math Is Fun Forum

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

please correct me

I know this can be done by drawing out vectors, but I'd like to know if my work is valid , Thanks in advance

Two balls A, B connected with a rigid stick L in length.  A is moving at a constant speed V ,when the angle is 60, what's the speed of B?

Last edited by Dragonshade (2008-06-16 11:29:41)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: please correct me

Maybe I'm being stupid here, but surely both balls need to be moving at the same speed?
If they had different speeds then they'd move relative to each other, which can't happen due to the rigid stick joining them together.

(Assuming there's no circular motion involved, because there's none mentioned in the question and if there was then the angle between the balls wouldn't be fixed.)

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: please correct me

Maybe this old post by Dross may help:
http://www.mathisfunforum.com/viewtopic … 090#p42090

Let a ladder of length L rest against a vertical wall. Let the horizontal distance from the base of the ladder to the base of the wall be given by x(t), and let the vertical distance between the base of the wall and the top of the ladder at time t be given by y(t).

Now, since the ladder, the distance x and the distance y form a right-angled triangle, we have by pythagoras that:

Now let the ladder be pulled away by it's base, such that the base moves away from the wall with a constant speed v. Differentiating the above with respect to t, we find that:

When the angle is 60°, substitute x = Lcos{60°) into the expression for

.

NB: It may look as if as x approaches L, B’s speed tends to infinity. This is not the case. In the case of the ladder, the top of the ladder will at some stage lose contact with the wall. In the case of this particular problem, since B cannot leave the horizontal axis, it will just mean that A will stop travelling with constant speed at the corresponding stage (so the above formula will then no longer apply).

Maybe I'm being stupid here …

Yes, you are being stupid here.

Last edited by JaneFairfax (2008-06-17 00:21:50)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: please correct me

Perhaps, but that doesn't help me figure out why.

The rigid stick and the fixed angle force the two balls to be stationary relative to each other, and so their speeds must match. Where is that wrong?

---

Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: please correct me

They are not stationary relative to each other. One is moving in the x direction and the other in the y direction. How can they be stationary relative to each other?

Last edited by JaneFairfax (2008-06-17 03:06:45)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: please correct me

Ohh, I see now. I interpreted that drawing very wrongly.

---

Why did the vector cross the road?
It wanted to be normal.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: please correct me

its my bad for not explaining it clearly
By vector, or function kinda of method (Jane mentioned)
I could work it out
I just wanna make sure, the 3rd method is valid, (I dont see any flaw)
Thanks~