Falling meter stick on a frictionless surface

In summary, the problem asks for the angular velocity of a meter stick that falls over. It is solved using equations and given that the end on the floor experiences no friction, the answer is w=sqrt(3g).
  • #1
harralk
18
0

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.
 
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  • #2
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.
 
  • #3
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?
 
  • #4
And what about h? Do I set it equal to the cm, 1/2m?
 
  • #5
harralk said:
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)
 
  • #6
ok, so substituting back in, I get w=sqrt(3g).

Thanks
 
  • #7
harralk said:
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 :redface: . Anyhow, it's still correct.
 
  • #8
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.
 

FAQ: Falling meter stick on a frictionless surface

What is a falling meter stick on a frictionless surface?

A falling meter stick on a frictionless surface is a common physics thought experiment that involves a meter stick being placed vertically on a surface with no friction, and then released from rest to fall under the influence of gravity.

What is the purpose of this thought experiment?

The purpose of this thought experiment is to understand the concepts of gravitational acceleration and the effect of friction on an object's motion.

What are the assumptions made in this thought experiment?

The main assumptions made in this thought experiment are that the surface is truly frictionless, there is no air resistance, and the meter stick is a rigid and uniform object.

What is the expected outcome of this thought experiment?

The expected outcome is that the meter stick will fall straight down without rotating or sliding, due to the absence of friction and air resistance.

How does this thought experiment relate to real-world scenarios?

This thought experiment is often used to illustrate the effects of gravity and friction in real-world scenarios, such as objects falling from a height or sliding down a ramp. It helps to understand the fundamental principles of motion and can be applied to various situations in physics.

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