Why is the result different in Method 2 for the rotating rod experiment?

In summary: This equation is very easy to understand once you understand that the parameter x is the distance between the rings and the center, and that the factor 0.9 works because the rings are moving away from the center at a speed of 0.9 meters per second.Thank you very much.
  • #1
Divya
9
1
Homework Statement
A rigid horizontal smooth rod $AB$ of mass $0.75 kg$ and length $40 cm$ can rotate freely about a fixed vertical axis through its mid point $O$. Two rings each of mass $1 Kg$ are initially at rest at a distance of $10 cm$ from $O$ on the either side of the rod.The rod is set in rotation with an angular velocity of $30$ radians per second. Find the velocity of each ring along the length of the rod in m/s when they reach the ends of the rod.
Relevant Equations
(I am new to this forum, I don't know what to write in 'Relevant Equations' field)
Method 1: Simply conserving angular momentum about the the fixed vertical axis and conserving energy gives ##v=3##, which is correct according to my book.

Method 2: Conserving angular momentum when the two rings reach distance ##x## from the centre gives
##(0.01+2x^2) \omega =0.9##
Also in the rod's frame ##a=v dv/dx =\omega ^2 x## (where a and v radial acceleration and velocities).
So, ##v^2/2=\int_{0.1}^{0.2} \frac {0.9xdx} {(0.01+2x^2)^2}=5##.
So, ##v=\sqrt{10}##.
What is wrong in second method?

Edit: Everything is in SI unit
 
Last edited:
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  • #2
The angular velocity of the rod decreases as the rings move outwards. The formula you used in the second method doesn't work with a reference frame that changes its angular velocity.

There are a lot of units missing.
 
  • #3
Why? The fictitious forces are centrifugal, coriolis and azimuthal. Here only centrifugal force is in radial direction. So, the equation should be correct.
 
  • #4
mfb said:
The angular velocity of the rod decreases as the rings move outwards. The formula you used in the second method doesn't work with a reference frame that changes its angular velocity.

There are a lot of units missing.
Sorry, everything is in SI unit.
 
  • #5
These are the forces in a reference frame where the angular velocity stays the same. I guess you could try to do that with the initial angular velocity but it just makes things needlessly complicated.
Divya said:
Sorry, everything is in SI unit.
And it should be added to the quantities that have units. It's a bad habit to not do that, especially combined with equations where quantities are not introduced first. 0.01 is what? Sure, it's possible to guess it from context, but it's just unnecessary. Don't let others guess.
 
  • #6
mfb said:
These are the forces in a reference frame where the angular velocity stays the same. I guess you could try to do that with the initial angular velocity but it just makes things needlessly complicated.And it should be added to the quantities that have units. It's a bad habit to not do that, especially combined with equations where quantities are not introduced first. 0.01 is what? Sure, it's possible to guess it from context, but it's just unnecessary. Don't let others guess.
As far I know, in case of non-uniformly rotating frame, everything is same except that we also need to add azimuthal force. And here azimuthal force is not in radial direction.
 
  • #7
Divya said:
##(0.01+2x^2) \omega =0.9##
Plugging in numbers straight away and with no text explanation makes it hard to follow the reasoning. Please explain the derivation of this equation. I was expecting ##x^2\omega=x_0^2\omega_0##.
 
  • #8
@haruspex, I have added the explanation of this equation in this attachment.
 

Attachments

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  • #9
Divya said:
@haruspex, I have added the explanation of this equation in this attachment.
Ok, thanks.
Your error is that you forgot to square the 0.9.
 
  • #10
haruspex said:
Ok, thanks.
Your error is that you forgot to square the 0.9.
I don't think so.
 
  • #11
Divya said:
##(0.01+2x^2) \omega =0.9##
So, ##v^2/2=\int_{0.1}^{0.2} \frac {0.9xdx} {(0.01+2x^2)^2}=5##.
## \omega =\frac {0.9}{0.01+2x^2}##
## \omega ^2=\frac {0.9^2}{(0.01+2x^2)^2}##
 
  • #12
Thank you very much.
 

Related to Why is the result different in Method 2 for the rotating rod experiment?

1. Why is Method 2 producing different results in the rotating rod experiment?

Method 2 may be producing different results because it utilizes a different approach or technique compared to Method 1. This could lead to variations in the data collected and ultimately affect the final result.

2. Could the equipment used in Method 2 be causing the difference in results?

It is possible that the equipment used in Method 2 may be a contributing factor to the difference in results. The instruments used in scientific experiments must be accurate and precise in order to obtain reliable data. Any discrepancies in the equipment could lead to variations in the results.

3. Are there any external factors that could be influencing the results in Method 2?

External factors such as temperature, humidity, or air currents could potentially affect the results in Method 2. It is important for scientists to control these variables in order to obtain accurate and consistent data.

4. Could human error be a factor in the difference between Method 1 and Method 2 results?

Human error is always a possibility in scientific experiments, and it could be a factor in the difference between Method 1 and Method 2 results. To minimize this, scientists must carefully follow the experimental procedure and take necessary precautions to ensure accuracy.

5. Is it possible that the difference in results between Method 1 and Method 2 is due to chance?

There is always a possibility that the difference in results between Method 1 and Method 2 is due to chance. In order to determine the significance of the difference, statistical analysis must be performed to determine the probability of obtaining the results by chance.

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