Easy question that I'm not getting?

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The discussion revolves around calculating the rotational inertia of a system consisting of a small ball mounted on a rod. The participant initially used the formula I = MR²/3, which is incorrect because the rod has negligible mass, making its inertia zero. The correct approach involves only considering the mass of the ball and using the formula I = MR², where R is the length of the rod. After realizing the mistake, the participant recalculates and arrives at an inertia value of 1.0115 for the system. This highlights the importance of correctly identifying which components contribute to rotational inertia in a system.
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Easy question that I'm not getting??

Homework Statement



A small ball with mass 1.40 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5020 rev/min.

Calculate the rotational inertia of the system about the axis of rotation.

Homework Equations



I=MR2/3

The Attempt at a Solution



Inertia for a rod spinning at one end = MR2/3, so I did that and got .337, but that wasn't the correct answer. What am I doing wrong?
 
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Come on, SOMEBODY has to know this =/
 


dominus96 said:

Homework Statement



A small ball with mass 1.40 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5020 rev/min.

Calculate the rotational inertia of the system about the axis of rotation.

Homework Equations



I=MR2/3

The Attempt at a Solution



Inertia for a rod spinning at one end = MR2/3, so I did that and got .337, but that wasn't the correct answer. What am I doing wrong?

The rod has (essentially) zero mass, so when you apply your equation you will find that the rotational inertia of the rod is zero.

But the system consists of more than the rod. What is the rotational inertia of this small ball about the axis of rotation?
 


Not quite sure what you mean. There is one rod and it has a mass attached to it, so don't you just use that as the mass in the inertia equation?
 


Wait, i think I get it. So do you do MR2/3 + MR2?
The inertia of the rod is 0, so the answer is just MR2? That would be 1.0115, am I right?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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