Rotational Dynamics Problem - Rod slipping against Block

Click For Summary
SUMMARY

The discussion revolves around a physics problem involving a uniform rod of mass m and length l, pivoted at point O, which loses contact with a block of mass M after being given a slight jerk. Key findings include the mass ratio M/m as 4:3, the block's velocity at the moment of contact loss calculated as √(3gl)/4, the center of mass acceleration of the rod at (3g)/4, and the hinge reaction force at O being (mg/4)j. The analysis emphasizes the importance of applying conservation of energy and understanding forces and accelerations to solve the problem accurately.

PREREQUISITES
  • Understanding of rotational dynamics and angular motion.
  • Familiarity with conservation of energy principles in mechanics.
  • Knowledge of forces acting on rigid bodies in motion.
  • Ability to differentiate and apply polar coordinates in dynamics.
NEXT STEPS
  • Study the application of conservation of energy in rotational systems.
  • Learn about the dynamics of rigid body motion, focusing on angular acceleration and forces.
  • Explore polar coordinate systems and their application in analyzing motion.
  • Investigate the relationship between linear and angular quantities in rotational dynamics.
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, as well as engineers and researchers dealing with rotational dynamics and contact forces in rigid body systems.

  • #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
 
Physics news on Phys.org
  • #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:

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 60 ·
3
Replies
60
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
11
Views
3K