Will a Meter Stick Rotate Around Its Center of Mass When Released?

AI Thread Summary
A uniform meter stick, when released from a horizontal position, will rotate around its center of mass, which is located at the 50cm mark. The only force acting on it post-release is gravity, which indeed acts at the center of mass, causing the rotation. The discussion also includes a question about the travel times of a mass projected along three different paths, with the same initial velocity. It is noted that the straight path (path 2) will result in the shortest travel time compared to the other paths. The conclusion emphasizes that the straight track allows for a faster arrival at point B due to the directness of the path.
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Very quick question: A uniform meter stick held by one end is swung in an arc and released when the person’s arm is horizontal, so that it moves initially away from the ground. About which point will it rotate as it flies before striking the ground?

Is it the 50cm mark since once released, the only force acting is gravity which acts at the centre of mass?

Thanks.
 
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Why would it rotate at all?
 
Whoops, you're right.

I have another question: A mass m starting at point A is projected with the same initial
horizontal velocity v0 along each of the three tracks shown here
(with negligible friction) sufficient in each case to allow the mass
to reach the end of the track at point B. (Path 1 is directed up,
path 2 is directed horizontal, and path 3 is directed down.) The
masses remain in contact with the tracks throughout their
motions. The displacement A to B is the same in each case, and
the total path length of path 1 and 3 are equal. If t1, t2, and t3 are
the total travel times between A and B for paths 1, 2, and 3,
respectively, what is the relation among these times?

Picture attached.

Does the ball in the straight track arrive faster? Why is that?
 

Attachments

These are the answer choices to the above question;

a) t3<t2<t1
b) t2<t3<t1
c) t2<t1=t3
d) t2=t3<t1
 
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