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**zetafunc.****Guest**

"One end of a rod of uniform density is attached to the ceiling in such a way that the rod can swing about freely with no resistance. The other end of the rod is held still so that it touches the ceiling as well. Then, the second end is released. If the length of the rod is *l* metres and gravitational acceleration is g metres per second squared, how fast is the unattached end of the rod moving when the rod is first vertical?"

I'm not sure I did this question correctly, but no answers are provided either. Here's what I did:

where I is the moment of inertia of the rod about an axis perpendicular to the rod and in the plane of the rod, and w is the angular velocity of the rod.

So we need to find the M.I. of the rod. Suppose I[sub]z[/sub] is the M.I. of the rod about an axis through one end of the rod and perpendicular to the rod. By the parallel axis theorem, the M.I. is

but we want I[sub]y[/sub]. But I[sub]x[/sub] = 0, so I[sub]y[/sub] = I[sub]z[/sub] by the perpendicular axis theorem. Since Loss in GPE = Gain in KE, then

therefore

is this correct?

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
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Phewh! For a second there I though it was our admin rotating.

The rod, it would seem represents a kind of pendulum whose period depends on uts mass, but the mass isn't given in the text, so I am a bit confused, and it doesn't look like a mathematical pendulum (where mass doesn't matter)...

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**zetafunc.****Guest**

I let the mass of the rod be m, but it appears that it doesn't matter, since the mass cancels when comparing the GPE and KE.

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
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Ok, step by step. How'd you get the loss in GPE formula?

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**zetafunc.****Guest**

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****Real Member**- From: Harlan's World
- Registered: 2011-05-23
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But, the question never states the starting position of the rod.

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**zetafunc.****Guest**

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****Real Member**- From: Harlan's World
- Registered: 2011-05-23
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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?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**zetafunc.****Guest**

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 ml[sup]2[/sup]/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****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,988

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

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
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**zetafunc.****Guest**

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****Moderator**- Registered: 2010-06-20
- Posts: 7,444

hi zetafunc

EDIT just read what you have asked. thinking ..........

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

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You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
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...... 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

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**zetafunc.****Guest**

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****Moderator**- Registered: 2010-06-20
- Posts: 7,444

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zetafunc.****Guest**

Okay, thank you!

**zetafunc.****Guest**

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

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