Angled falling Meter Stick on a frictionless surface

[stormfalconx]

Homework Statement

A meter stick is held 60 degrees to the horizontal on a flat frictionless surface. If it is let go what is the distance the back the stick will reach with respect to the length of the meter stick (l).

Homework Equations

I = 1/3 mL^2
Don't know what to use. This seems more conceptual.

The Attempt at a Solution

I assumed that the bottom of where the meter stick touches the ground is 0. If the stick is held at an angle and there is no net torque in the x-direction, then the center of mass should fall straight down right? I don't know where to go from there. How would I find out how far back the bottom of the meter stick has moved?

 

[haruspex]

How would I find out how far back the bottom of the meter stick has moved?

Consider horizontal forces.

 

[stormfalconx]

Consider horizontal forces.

Do you mean the torque that is opposite the direction of falling? How does torque relate to distance though? I thought it was only equal to I*α or rFsinθ?

 

[haruspex]

Do you mean the torque that is opposite the direction of falling? How does torque relate to distance though? I thought it was only equal to I*α or rFsinθ?

No, not torque. Just the horizontal forces and any consequential acceleration.

 

[stormfalconx]

No, not torque. Just the horizontal forces and any consequential acceleration.

How can there be a horizontal force when an object is dropped? Am I missing a component of Fg somewhere?

 

[haruspex]

How can there be a horizontal force when an object is dropped? Am I missing a component of Fg somewhere?

Right, there is no horizontal component of Fg. Any other possibility?

 

[kuruman]

... then the center of mass should fall straight down right? I don't know where to go from there. How would I find out how far back the bottom of the meter stick has moved?

If the center of mass falls straight down, can the low end of the stick remain in its starting position? Draw a "before" and "after" picture.

 

[haruspex]

If the center of mass falls straight down, can the low end of the stick remain in its starting position? Draw a "before" and "after" picture.

Thanks for stepping in. I had fixated on the OP's reference to torque and missed that the OP had already figured out that the mass centre will descend vertically.

 

[stormfalconx]

If the center of mass falls straight down, can the low end of the stick remain in its starting position? Draw a "before" and "after" picture.

I think the low end of the stick will move back because the surface is frictionless. I'm attaching a drawing I made. I made the red dot the center.

 

[stormfalconx]

Thanks for stepping in. I had fixated on the OP's reference to torque and missed that the OP had already figured out that the mass centre will descend vertically.

I attached an image on post 9. I know the center of mass will stay vertically the same. However I can't wrap my head around how far the stick will go based on just this information.

 

[haruspex]

I attached an image on post 9. I know the center of mass will stay vertically the same. However I can't wrap my head around how far the stick will go based on just this information.

It will help to put both positions in the one diagram, and mark the start and finish positions of the centre of the stick. Show all known distances and angles, and the distance to be found as x.

 

[Dr.D]

Consider horizontal forces.

What horizontal forces did you have in mind on a frictionless table top?

 

[haruspex]

What horizontal forces did you have in mind on a frictionless table top?

That was my point, but at that stage I had not noticed the OP had already figured out the mass centre would fall vertically.

 

[stormfalconx]

It will help to put both positions in the one diagram, and mark the start and finish positions of the centre of the stick. Show all known distances and angles, and the distance to be found as x.

Thanks I think figured it out. When I drew it out I solved for horizontal distance between the CM and point of contact with the floor. That was 0.25m and so if the CM falls straight down, it will cover 0.25m of its 0.5m for the initial horizontal distance, leaving another 0.25 meters extending over the initial point of contact.

 

[haruspex]

Thanks I think figured it out. When I drew it out I solved for horizontal distance between the CM and point of contact with the floor. That was 0.25m and so if the CM falls straight down, it will cover 0.25m of its 0.5m for the initial horizontal distance, leaving another 0.25 meters extending over the initial point of contact.

Right!