Recent content by stormfalconx

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...

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.

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.

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

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?

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...