Conceptual questions about rotational dynamics.

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The discussion centers on a problem involving a dumbbell with two unequal masses rotating about a point, focusing on how to double its rotational kinetic energy while keeping angular velocity constant. The initial conclusion was that the distance R must increase by a factor of two, but the correct factor is the square root of two, as derived from the relationship that kinetic energy is proportional to R squared. A conceptual question about moment of inertia clarified that it exists regardless of whether an object is rotating, as it depends on the axis of rotation. The preferred method for solving such problems involves setting up equations based on kinetic energy and systematically solving for the variables, rather than relying solely on numerical substitution. Understanding the proportionality of kinetic energy to R squared is crucial for accurate problem-solving in rotational dynamics.
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Homework Statement


There is a dumbbell with two masses attached at each end, rotating about a point. The distance between the two masses is R, and the two masses do not have the same mass. m1 < m2

If the angular velocity ω is held constant, by what factor must R change to double the rotational kinetic energy of the dumbbell?

Homework Equations


I = mr^{2}
r = \frac{R}{2}
K_{rot} = \frac{1}{2}I\omega^{2}

The Attempt at a Solution


Since r = \frac{R}{2} and I = (m_{1}+m_{2})r^{2}, then K_{rot} = \frac{1}{2}[(m_{1}+m_{2})((\frac{R}{2})^{2})]\omega^{2} =&gt;\frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{4})]\omega^{2}

If Rotational kinetic energy is doubled and angular velocity is to be constant, then:
2K_{rot} = 2(\frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{4})]\omega^{2})
2K_{rot} = \frac{1}{2}[(m_{1}+m_{2})(\frac{2R^{2}}{4})]\omega^{2}
2K_{rot} = \frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{2})]\omega^{2})

Thus, by my conclusion, R increases by a factor of two if the rotational energy is doubled but angular velocity is to remain constant. My book's answer is that R increases by a factor of square root of 2. What did I do wrong?

Also, one more conceptual question: If an object is not rotating, does it have a moment of inertia? The answer is no, correct? Since a moment of inertia depends on the rotation of axis and changes when an object is rotating about a different point, thus if an object is not rotating, then it does not have a moment of inertia since moment of inertia is the rotational equivalent of mass.
 
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Retribution said:
Thus, by my conclusion, R increases by a factor of two if the rotational energy is doubled but angular velocity is to remain constant. My book's answer is that R increases by a factor of square root of 2. What did I do wrong?
Try this:
KE1 = f(R1)
KE2 = f(R2)
Set KE2 = 2KE1 and then solve for R2 in terms of R1.

Also, one more conceptual question: If an object is not rotating, does it have a moment of inertia? The answer is no, correct? Since a moment of inertia depends on the rotation of axis and changes when an object is rotating about a different point, thus if an object is not rotating, then it does not have a moment of inertia since moment of inertia is the rotational equivalent of mass.
If an object isn't moving does it still have mass? Same idea here. Of course it has a moment of inertia (which depends on the axis you choose to rotate about).
 
Doc Al said:
Try this:
KE1 = f(R1)
KE2 = f(R2)
Set KE2 = 2KE1 and then solve for R2 in terms of R1.
The major mistake I made was to assume that R denoted the length, not the radius. In that case:
\frac{1}{2}(m_{1} + m_{2})(R_{2})^{2}\omega^{2} =2[\frac{1}{2}(m_{1} + m_{2})(R_{1})^{2}\omega^{2}]
Masses and angular velocity cancels out:
\frac{1}{2}(R_{2})^{2} =2[\frac{1}{2}(R_{1})^{2}]
R_{2} = \sqrt{2R_{1}}

Thanks for all the help. I just wanted to know, is this the preferred method of going about solving these types of questions? I would usually plug in numbers, but that method mislead me into thinking I had the right answer.
Doc Al said:
If an object isn't moving does it still have mass? Same idea here. Of course it has a moment of inertia (which depends on the axis you choose to rotate about).
Ah, I see. Thanks.
 
Retribution said:
Thanks for all the help. I just wanted to know, is this the preferred method of going about solving these types of questions? I would usually plug in numbers, but that method mislead me into thinking I had the right answer.
It's how I would approach a problem like this. The main thing is to realize that the KE is proportional to R2. So to double the KE you need to multiply R by √2.

But doing it systematically reduces the chance for error. (And may lead to a deeper understanding.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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