Rotational Dynamics Problem - Rod slipping against Block

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Homework Help Overview

The problem involves a uniform rod pivoted at one end, initially vertical, which begins to rotate after being given a slight jerk. The rod interacts with a block on a horizontal surface, and the discussion focuses on the conditions under which the rod loses contact with the block as it rotates. Participants are exploring concepts in rotational dynamics, energy conservation, and forces acting on the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the speed of the block and the tip of the rod, questioning the correctness of their initial assumptions about angular and linear speeds.
  • There is exploration of energy conservation principles and the need to incorporate forces and accelerations into the analysis.
  • Some participants express uncertainty about how to relate the forces acting on the rod and block, particularly when the normal force disappears.
  • Questions arise regarding the use of polar coordinates to describe the motion of the rod's tip and how to derive the acceleration components.

Discussion Status

The discussion is ongoing, with participants actively questioning their approaches and seeking clarification on the relationships between various physical quantities. Some have identified mistakes in their reasoning and are attempting to correct them, while others are exploring different methods to analyze the problem. There is no explicit consensus yet, but productive dialogue is occurring around the concepts of forces, torques, and energy conservation.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for analysis. The problem setup involves smooth surfaces and specific angles at which contact is lost, which are critical to the discussion but not fully resolved.

  • #31
The normal force on the block is positive or zero. ω2 is positive, but the angular acceleration can be negative.

Have you got the direction of the acceleration of the COM?

ehild
 
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  • #32
ehild said:
Have you got the direction of the acceleration of the COM?

ehild

Yes , I have got the direction of acceleration of COM.I worked with radial and tangential components and found their resultant i.e acom to make 30° with the tangential component .i.e vertically downwards .This showed that the net acceleration of COM was vertically downwards at the moment rod loses contact with the block .The net resultant force should be vertically downwards i.e -ve y-axis .Now N vanishes so horizontal component of Hinge force is zero . :smile:
 
  • #33
Very well! So you got the correct result that the reaction force is mg/4 at the hinge upward.

You would get the same from the x,y components of acceleration. I like this method, as I usually get confused with the radial and tangential components.
The coordinates of the COM of a homogeneous rod with the ends at (x1,y1) and (x2,y2) are Xcom=(x1+x2)/2, Ycom=(y1+y2)/2. (x1,y1)=(0,0), and (x2,y2)=(x,y) here, so Xcom=x/2, Ycom=y/2.

You derived the x component of acceleration already, the same has to be done with the y component.

y=Lsin(θ) y'=Lcos(θ) θ',

y''= L(-sin(θ)θ'2+cos(θ)θ")=L(-sinθω2+cos(θ)ω')

When loosing contact, y"=L(-1/2 ω2+√3/2 ω')

From the condition that the x component of acceleration is zero, you have got that

ω'=-√3 ω2 and ω2=3g/4L, therefore y"=-3g/2,

and Ycom"=-3g/4.

ehild
 
  • #34
ehild...Thank you very much ...

Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question :biggrin:
 
  • #35
Tanya Sharma said:
ehild...Thank you very much ...

Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question :biggrin:

So you discovered that it was about relation between polar and Cartesian coordinates? I did not want to confuse you by speaking about polar coordinates: :-p

ehild
 
  • #36
yes process is correct
 
Last edited:

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