What is the Time and Distance for a Meter Stick to Fall?

[guss]

I've been thinking about this simple problem a lot lately and it has been bugging me.

Let's say you have a meter stick that weighs 100g and is 1x.001x.001 meters (so it's more like a rod). You probably don't need all that information but I'll include it to be safe. The meter stick is balancing straight up so the top of the stick is a meter off the ground.

You push the top of the meter stick just enough. So it begins to tip in one direction (let's assume you tip it in the direction of one of the sides of the stick).

a) how long will it take the meter stick to fall if the bottom of the stick does not slide on the ground?

b) if the coefficients of friction between the stick and the ground is .1, how long will the stick take to fall (I think it's the same as part a)? How far will the center of the stick be from where it was when it landed in part a? After the stick lands, will it slide along the ground at all? (I don't think it will)

Thanks.

 

[tiny-tim]

hi guss!

a) how long will it take the meter stick to fall if the bottom of the stick does not slide on the ground?

b) if the coefficients of friction between the stick and the ground is .1, how long will the stick take to fall (I think it's the same as part a)? How far will the center of the stick be from where it was when it landed in part a? After the stick lands, will it slide along the ground at all? (I don't think it will)

a) use conservation of energy

b) find the https://www.physicsforums.com/library.php?do=view_item&itemid=73" in the a) situation … that will tell you when the stick will start to slide

 

[guss]

I'm still confused, not sure where to start.

 

[Redbelly98]

Has your class discussed conservation of energy yet?

 

[guss]

Has your class discussed conservation of energy yet?

Of course. I'm just not sure how to work that into this problem. I haven't done much calculus based physics, but I've done calculus.

 

[nasu]

You can find the time from conservation of energy?

@guss
Is this a problem that you made up yourself?
For the first case maybe you need to consider a bar articulated at the bottom end.

 

[guss]

You can find the time from conservation of energy?

That's what I was trying to figure out. Still not quite sure how they are going to use conservation of energy to do that

Is this a problem that you made up yourself?

Yes.

 

[ngc1333]

That's what I was trying to figure out. Still not quite sure how they are going to use conservation of energy to do that


Yes.

Do you know calculus? I swear I posted something here last night, but it's gone now. If you know calculus I'll post it again. If not, I don't see how you can solve the problem.

 

[guss]

Do you know calculus? I swear I posted something here last night, but it's gone now. If you know calculus I'll post it again. If not, I don't see how you can solve the problem.

I know a good amount of calculus. I actually found this (from you) in my email:

I = \frac{1}{3}ml2

The initial energy of the rod is

E0 = mg\frac{l}{2}.

At any later time, the energy is

E = \frac{1}{2}mgycm+\frac{1}{2}I\omega2

From conservation of energy:

mg\frac{l}{2} = \frac{1}{2}mgycm+\frac{1}{2}I\omega2

The y coordinate of the center of mass is ycm=l sin \theta. Plugging that in, and solving for \omega gives:

\omega = [\frac{3g}{l}(1-sin\theta)]-1/2

Using the definition \omega=\frac{d\theta}{dt}, we can write

[\frac{3g}{l}(1-sin\theta)]-1/2d\theta=dt

You can then integrate over \theta\in(\frac{\pi}{2},0) to find the time.
- Show quoted text -

So I guess that helps a bit ;)

 

[ngc1333]

Interesting. I wonder what happened to that post. I tried to edit a mistake (the line after the "From conservation of energy" has an extra 1/2 that I tried to get rid of). I guess I screwed something up editing it.

 

[Redbelly98]

I swear I posted something here last night, but it's gone now.

ngc1333, you should have received a Private Message (PM) from me about that. Did you not receive it? Please check your Private Messages.

 

[ngc1333]

ngc1333, you should have received a Private Message (PM) from me about that. Did you not receive it? Please check your Private Messages.

Got it.