Moment of Inertia and Angular Momentum

AI Thread Summary
The discussion focuses on calculating the speed of the center of mass of a thin meter stick just before it hits a frictionless table. The correct approach involves using the parallel axis theorem to find the moment of inertia since the stick rotates about one end, not its center. The initial calculations provided by one participant yielded a speed of 3.13 m/s, while the answer key states it should be 2.71 m/s. The gravitational potential energy at the center of the stick is calculated to be 0.981 J, which is then equated to the rotational kinetic energy. Clarification on the use of the parallel axis theorem is emphasized to resolve the discrepancy in the answers.
Kommandant
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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
 
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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.
 
Could you show me step by step how to do it
 
Kommandant said:
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?
 
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
 
Kommandant said:
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=Ic+mr2
 
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