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A physics professor that doesn’t get easily embarrassed stands at the center of a frictionless turntable with arms outstretched and a 5.0-kg dumbbell in each hand. He is set rotating about the vertical axis, making one revolution in 2.0s. Find his angular velocity if he pulls the dumbbells to his stomach. His moment of inertia (without the dumbbells) is 3.0 kg*m^2 with arms outstretched and 2.2 kg*m^2 with his hands at his stomach. The dumbbells are 1.0 m from the axis initially and 0.20 m at the end. How about Kinetic Energy before and after? Explain from where this energy (if any) came from?

Respuesta :

rokettojanpu
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.

L must stay the same before and after the professor brings the dumbbells closer to himself.

His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2

His final moment of inertia is 2.2kg•m^2.

Calculate the initial angular velocity:
L = 3.0π

Final angular velocity:
L = 2.2w

Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:

3.0π = 2.2w
w = 1.4π rad/s

The rotational energy is given by:
KE = 0.5Iw^2

Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J

Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J

There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
migueletorallam

Answer:

[tex]\omega_{f} = 5\pi[/tex]

Energy comes from the work the professor does to move the dumbbells towards him.

Explanation:

To find the angular velocity of the professor we are going to use the conservation of angular momentum, this is [tex]L_{0}=L_{f}[/tex].

At the begining we have that:

[tex]L_{0} =\omega_{0}I_{0}[/tex] .

We can easily find [tex]\omega_{0}[/tex] because we know the oscilation period (T). So:

[tex]\omega_{0}=\frac{2\pi}{T}[/tex],

[tex]\omega_{0}=\frac{2\pi}{2}[/tex],

[tex]\omega_{0}=\pi(rad/s)[/tex].

The moment of inertia at the beginning is the sum of his moment of inertia and the moment of inertia of dumbbells(consider the dumbbells as particles so their moment of inertia is [tex]mr^{2}[/tex], where m is the mass and r is the distance from the dumbbell to the axis), so:

[tex]I_{0}=2mr^{2}+3[/tex]

[tex]I_{0}=2(5)(1)^{2}+3[/tex]

[tex]I_{0}=13(kg*m^2)[/tex].

The moment of inertia at the final position is meant to be computed the same way (using the values of the ending position) :

[tex]I_{f}=2mr^{2}+2.2[/tex]

[tex]I_{f}=2(5)(0.2)^{2}+2.2[/tex]

[tex]I_{f}=2.6(kg*m^2)[/tex].

Now we recall the conservation of angular momentum to compute his final angular velocity:

[tex]L_{0}=L_{f}[/tex]

[tex]I_{0}\omega_{0}=I_{f}\omega_{f}[/tex]

[tex]\omega_{f}=\frac{I_{0}\omega_{0}}{I_{f}}[/tex]

[tex]\omega_{f}=\frac{(13)(\pi)}{2.6}[/tex]

[tex]\omega_{f}=5\pi(rad/s)[/tex].

If we look at the rotational kinetic energy ([tex]K_{rotational}=\frac{1}{2}I(\omega)^{2}[/tex]) we realized that the energy at the start (64.15 J) is less than the one at the end (320.76 J). This is due to the work the professor did to move the dumbbells toward him.

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