Finding Angular Velocity and Tip Speed of a Falling Rod

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SUMMARY

The discussion focuses on calculating the angular velocity and tip speed of a falling rod, specifically a thin rod of length L that pivots about its lower end. The angular velocity as a function of the angle φ is derived using the equation α(φ) = (3g sin φ) / (2L), leading to the conclusion that ω = (3g / 2L) rad/sec. The importance of the condition "falls without slipping" is emphasized, indicating that the rod pivots at its lower end, which is crucial for correctly applying the moment of inertia formula I = (1/3)mL².

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with moment of inertia calculations for rigid bodies
  • Basic knowledge of angular acceleration and its relationship to torque
  • Concept of pivot points in rigid body motion
NEXT STEPS
  • Study the derivation of angular velocity in rotational motion problems
  • Explore the concept of moment of inertia for various shapes and configurations
  • Learn about the effects of friction in pivoting systems
  • Investigate the dynamics of falling objects and their angular motion
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of rotating bodies and angular motion.

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Homework Statement



A thin rod of length L stands vertically on a table. The rod begins to fall, but its lower end does no slide. (a) Determine the angular velocity of the rod as a function of the angle \phi it makes with the tabletop. (b) What is the speed of the tip of the rod just before it strikes the table?


Homework Equations



1) Torque = Moment of Inertia * angular acceleration
T = I * \alpha

T = Fdsin\phi



The Attempt at a Solution



I am not sure if this is the correct solution because I'm not sure if I'm thinking about the problem correctly.

We know there is a force from gravity Fg. It will depend on the angle of the rod with the table. This relation is given by Fg(\phi) = mgsin\phi.

so T(\phi) = mgsin\phid = mg(L/2)sin\phi.

now \alpha = \frac{T}{I}, so

\alpha(\phi) = (mgLsin\phi)/2I,

and we know I for a rod in this case is 1/3mL^2

so \alpha(\phi) = \frac{3gsin\phi}{2L}])

then I integrate the ang. acceleration to find the angular velocity from \pi/2 to 0.

I found \omega = \frac{3g}{2L} rad/sec .


I have good feeling I am leaving something out in this solution, such as possibly something to do with friction with the surface. I have a feeling the words "falls without slipping" mean something I am not realizing.
 
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I have a feeling the words "falls without slipping" mean something I am not realizing.
No, one correctly interpreted the "falls without slipping", which means that the rod pivots about the lower end, rather than at the center of gravity or point in between. One selected the appropriate from of the MI for this condition.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod
 
Ah, yes. Sometimes when thinking about more complex problems I start getting confused about the simple things and start questioning everything I normally know.

Thanks anyways.
 

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