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

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SUMMARY

A uniform meter stick, when released from a horizontal position, rotates around its center of mass, which is located at the 50 cm mark. The only force acting on the stick after release is gravity, which acts at the center of mass, causing it to rotate as it falls. Additionally, a mass projected along three different tracks (upward, horizontal, and downward) will have varying travel times, with the straight horizontal path being the fastest due to the lack of vertical displacement affecting the time of travel.

PREREQUISITES
  • Understanding of center of mass and rotational motion
  • Basic principles of projectile motion
  • Knowledge of gravitational forces and their effects on objects
  • Familiarity with kinematics and time-distance relationships
NEXT STEPS
  • Study the principles of rotational dynamics and moment of inertia
  • Learn about projectile motion equations and their applications
  • Explore the concept of energy conservation in different motion paths
  • Investigate the effects of friction on motion in various scenarios
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Physics students, educators, and anyone interested in understanding the mechanics of motion and forces acting on objects in free fall.

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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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