Rotational Dynamics Problem - Rod slipping against Block

  • #26
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acom=3g/4.I have calculated it from the radial and tangential components.

Is there another way of calculating it??
 
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  • #27
ehild
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What is the direction of acom?
You can get the components of the acceleration of COM also from the acceleration of the tip.
acomx=0.5d2x/dt2, acomy=0.5d2y/dt2

ehild
 
  • #28
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How are both the x and y components half of the acceleration of the tip of the rod?? I mean , I dont understand how the accelerations of COM and the tip are related ??
 
  • #29
ehild
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How are the coordinates of the COM related to those of the end of rod, supposed that the origin is at the hinge?

ehild
 
  • #30
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Okay...i understand what you say...

Why is there a -ve sign in the expression for N and ω2 ? I mean what does the -ve sign signify considering ω2 is a positive quantity.
 
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  • #31
ehild
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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
 
  • #32
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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
ehild
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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
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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
ehild
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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: :tongue2:

ehild
 
  • #36
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yes process is correct
 
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