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zetafunc.
2012-10-12 07:32:37

A maths teacher said I am wrong but never explained why. Confused...

zetafunc.
2012-10-11 07:29:12

Okay, thank you!

bob bundy
2012-10-11 05:23:52

Oh yes, I'm terrible for only half reading the question.  It does ask 'how fast' so I guess they want the velocity.

As you've got omega first I think you've covered yourself whatever.

Bob

zetafunc.
2012-10-11 02:57:39

I've studied moments of inertia before (had to do it for my exam) and deriving them via integration -- but I know the M.I. of the rod is definitely correct (and therefore the change in KE). Since:

as v = rw. But, this question still confuses me... are they asking for the ANGULAR velocity w, or just the instantaneous velocity of the end of the rod? In which case, with r = l;

v = rw, using my w from post #1;

bob bundy
2012-10-10 18:24:49

...... stopped thinking

I suppose you could find a way of calculating the KE directly.  Write the velocity in terms of omega and l.

Maybe try it.

B

bob bundy
2012-10-10 18:21:40

hi zetafunc

Moments of inertia can be calculated by integration.  It's ages since I did this so there's lots of cobwebs to blow away first.

I'm sure you could do it.

MI of whole = integral over length of rod (MI of a dx segment)

Bob

zetafunc.
2012-10-10 18:09:22

Thanks.

Is there a way to do this without knowing anything about moments of inertia? This question is taken from an admissions test in 2007 from Trinity College, Cambridge, but most applicants typically will not have covered this until Year 13 (it's an M5 topic)... can it be done with only M1-M3 knowledge?

bob bundy
2012-10-10 18:03:29

hi zetafunc,

That looks good to me.

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Bob

anonimnystefy
2012-10-10 10:23:06

I would have to look at this in the morning. My brain is not functioning properly right now.

zetafunc.
2012-10-10 10:13:29

For a uniform rod of mass m and length 2l, the moment of inertia about an axis perp. to the rod through the centre is ml2/3 (you can prove this by integration but it should be fine to quote it). For a rod of length l, you square (l/2) instead. Then apply the parallel axis theorem, to get the M.I. of the rod about the axis through the end. I sort of wasted time thinking about I_x, I_y and I_z, since it is a rod, not a lamina, and therefore I_y = I_z.

anonimnystefy
2012-10-10 10:07:08

Oops! Me and my careful reading. Didn't see the "also touches the wall" part. Sorry.

It looks like the mass does cancel out, after all.

Just one more thing-Why are you using l/2 in tbe I_z formula if the rotation is around the end of the rod?

zetafunc.
2012-10-10 09:54:03

One end of a rod is attached to the ceiling in such a way that the rod can swing about freely. The other end of the rod is held still so that it touches the ceiling as well. Then, the second end is released.

I don't see how else this could be interpreted... clearly the starting position is the rod in its horizontal position.

= (left end fixed)

goes to

|| (top end fixed)

anonimnystefy
2012-10-10 09:49:36

But, the question never states the starting position of the rod.

zetafunc.
2012-10-10 09:42:37

If the rod is at first horizontal, then rotates 90 degrees (when first vertical), then the change in GPE is mgl/2, since the centre of mass of the rod has dropped by a distance of l/2.

anonimnystefy
2012-10-10 09:40:18

Ok, step by step. How'd you get the loss in GPE formula?