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Moment of inertia Question

  • Thread starter Lord Dark
  • Start date
  • #1
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Homework Statement


hello everyone ,, can someone tell me if I got the moment of inertia right or wrong of this question :
Two blocks of masses m1 = 5 kg and m2 = 15 kg are suspended from the ends of 1.5 m
rigid rod of weight 75 N that can rotate about point P, as shown in the figure. The rod is
held in a configuration such that it makes an angle of 37° with the vertical, and then
released. The two blocks can be considered as point particles and the moment of inertia
of the rod about its center of mass is I(com) = ML^2/12.


Homework Equations





The Attempt at a Solution


I=I(com)+M(L/3)^2+m1(2L/3)^2+m2(L/3)^2 = (7/36)ML^2+(m2L^2/9)+(4m1L^2/9),, my teacher told me that adding m2 and m1 is wrong ,, but what i don't get ,, in some questions we add them ,, so is he right or wrong ??
 

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Answers and Replies

  • #2
LowlyPion
Homework Helper
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And the figure shows what exactly?
 
  • #3
Doc Al
Mentor
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Attach the figure.
 
  • #4
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sorry ,, forgot to add it
 
  • #5
Doc Al
Mentor
44,877
1,129
The masses are not rigidly attached to the stick, so don't include them when calculating the rotational inertia.
 
  • #6
121
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how come we don't include them ?? in the previous question (https://www.physicsforums.com/showthread.php?t=314630) we added them to the moment of inertia ,, in Fundamentals of physics book there is a question in Rotation chapter (11) Q:57P ,, in the answer .. it I*Alfa = (mL^2(1)+ mL^2(2))*alfa and it's almost the same as the question but without angle (it's horizontal) but I equaled : (mL^2(1)+ mL^2(2)) (the rod is mass less in this question)
By rigidly ,, you mean it fixed and cant be taken away ? ,, BTW some questions say (without the cord slipping on the pulley) does it mean it fixed ?
 
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