Moment of Inertia and Angular Momentum

[Kommandant]

Homework Statement

This was a TMSCA (Texas Math and Science Competition) from the physics section.

A thin meter stick of mass 200g is positioned vertically on a frictionless table. It is released, slips, and falls. Find the speed of its center of mass just before it hits the table.

Homework Equations

To the right of this question there is the equation for the Moment of Inertia of the stick
I=(1/12)mL^2

The Attempt at a Solution

Answer seems to be 2.71 according to the Key
Answer that I got was 3.13 m/s

 

[rock.freak667]

The stick is rotating about one end, not its center, so use the parallel axis theorem to get the moment of inertia about one end. That should give you the correct answer.

 

[Kommandant]

Could you show me step by step how to do it

 

[rock.freak667]

Could you show me step by step how to do it

Post what you did. I am guessing your problem other than the moment of inertia?

 

[Kommandant]

I found the gravitational potential energy at the center of the meter stick (.5 meters)

PEg= (.2)(9.81).5) = .981

Then I made .981 equal to the Kinetic Energy of an object in rotation

.981= (1/2)Iw^2

Where I is Moment of Inertia and w is angular velocity
I=MR^2
1.962=(.2)(.5^2)w^2
w= 6.2641

Then V= WR
V= 6.2641 * (1/2)
V=3.13 m/s

 

[rock.freak667]

I found the gravitational potential energy at the center of the meter stick (.5 meters)

PEg= (.2)(9.81).5) = .981

Then I made .981 equal to the Kinetic Energy of an object in rotation

.981= (1/2)Iw^2

Where I is Moment of Inertia and w is angular velocity
I=MR^2
1.962=(.2)(.5^2)w^2
w= 6.2641

Then V= WR
V= 6.2641 * (1/2)
V=3.13 m/s

your error is there, for a rod, I=(1/12)ML² about its center. But it is not rotating about its center, it is rotating about its end. Thus you need to use the parallel axis theorem. Do you know the parallel axis theorem?

I=I_c+mr²