Rotational Dynamics Problem - Rod slipping against Block

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A uniform rod pivoted at point O rotates after being jerked, causing a block to move forward until the rod loses contact at an angle of 30°. The mass ratio of the block to the rod is determined to be 4:3, with the block's velocity at separation calculated as √(3gl)/4. The acceleration of the rod's center of mass at this point is (3g)/4, and the hinge reaction force is mg/4. The discussion emphasizes the importance of using forces and accelerations to analyze the system, particularly when determining the moment of contact loss. Understanding the relationship between angular and linear motion is crucial for solving the problem effectively.
  • #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
 
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