Falling meter stick on a frictionless surface

[harralk]

Homework Statement

The question I'm working on says "A special, frictionless, thin meter stick is initially standing vertically on the floor. If the meter stick falls over, with what angular velocity will it hit the floor?"

Homework Equations

v_rot=rw, I=1/12mL^2, KE=1/2Iw^2

The Attempt at a Solution

I know that the stick is going to be spinning about its center of mass, assumably in the center of the stick. I also know that the cm will be falling straight down. When I set mgh=1/2Iw^2 I get stuck trying to solve for w without being given the mass of the stick.

 

[rock.freak667]

The mass of the stick as at the geometric centre. The point where it touches the ground, that acts like a pivot. So yes you must use mgh=1/2 Iω²

But the 'I' in this case is not about the centre but about one end.

 

[harralk]

Thanks. But I forgot to add that the problem also says "assume that the end in contact with the floor experiences no friction and slips freely." I assumed they meant that the end on the floor would be moving in the opposite direction and with equal velocity as the top of the stick, hence the rotation about the cm. Do I need to use I=1/3mL^2?

 

[harralk]

And what about h? Do I set it equal to the cm, 1/2m?

 

[rock.freak667]

Thanks. But I forgot to add that the problem also says "assume that the end in contact with the floor experiences no friction and slips freely." I assumed they meant that the end on the floor would be moving in the opposite direction and with equal velocity as the top of the stick, hence the rotation about the cm. Do I need to use I=1/3mL^2?

Well it still means it rotates about the end, slips freely and no friction means that in your energy equation, you don't need to account for friction.

So yes I=(1/3)mL²

so what in mgh=(1/2)Iω², what does I change to? and what is h in terms of L? (the centre of mass is at the geometric centre of the stick)

 

[harralk]

ok, so substituting back in, I get w=sqrt(3g).

Thanks

 

[rock.freak667]

ok, so substituting back in, I get w=sqrt(3g).

Thanks

Yep...I was almost going to tell you it should be ω=√(3g/L) when I remembered it was a metre stick . Anyhow, it's still correct.

 

[harralk]

FYI...In case anyone is reading this thread for help with their own homework, I thought about the problem one last time before I turned it in and changed my answer. After receiving my homework back, with the correct answer, the actual formula would be sqrt((24gh)/(l^2)). The way to think about it is since the stick is standing on a frictionless surface, once the top of the stick moves past vertical there is an x-component of its weight pushing the bottom sideways. Therefore, the stick rotates about its center (I=1/12MR^2) with the cm falling straight down. Be careful, because this problem didn't indicate an initial velocity for the top of the stick, which means that it was allowed to just fall over not pushed over.

Hope this helps clarify the wonderful world of physics.