A physics professor that doesn’t get easily embarrassed stands at the center of a frictionless turntable with arms outstretched

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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?

Physics

2 answers:

Ostrovityanka [3.2K] · Answer 1 · 2026-03-10T07:40:17+03:00

Answer:

$\omega_{f} = 5\pi$

The energy originates from the effort the professor exerts to pull the dumbbells closer.

Explanation:

We will utilize the conservation of angular momentum to determine the professor's angular velocity, illustrated in $L_{0}=L_{f}$ .

Initially, we can establish that:

$L_{0} =\omega_{0}I_{0}$ .

We can determine $\omega_{0}$ easily since we have the period of oscillation (T). Therefore:

$\omega_{0}=\frac{2\pi}{T}$ ,

$\omega_{0}=\frac{2\pi}{2}$ ,

$\omega_{0}=\pi(rad/s)$ .

The initial moment of inertia combines his own moment of inertia and that of the dumbbells (considering the dumbbells as point masses, their moment of inertia is $mr^{2}$ , where m refers to mass and r is the distance from the axis), thus:

$I_{0}=2mr^{2}+3$

$I_{0}=2(5)(1)^{2}+3$

$I_{0}=13(kg*m^2)$ .

To find the moment of inertia at the final position, we compute it in the same manner (using the final position values):

$I_{f}=2mr^{2}+2.2$

$I_{f}=2(5)(0.2)^{2}+2.2$

$I_{f}=2.6(kg*m^2)$ .

Now we apply the conservation of angular momentum to calculate his final angular velocity:

$L_{0}=L_{f}$

$I_{0}\omega_{0}=I_{f}\omega_{f}$

$\omega_{f}=\frac{I_{0}\omega_{0}}{I_{f}}$

$\omega_{f}=\frac{(13)(\pi)}{2.6}$

$\omega_{f}=5\pi(rad/s)$ .

Upon examining the rotational kinetic energy ( $K_{rotational}=\frac{1}{2}I(\omega)^{2}$ ), we find that the initial energy (64.15 J) is lower than the final energy (320.76 J). This increase is attributed to the work done by the professor in moving the dumbbells inward.

Sav [3.1K] · Answer 2 · 2026-03-10T07:41:45+03:00

To apply the conservation of angular momentum:
L = Iw = constant.
Here, L stands for angular momentum, I denotes moment of inertia, and w signifies angular velocity; L remains constant.

Angular momentum prior and subsequent must remain unchanged when the professor brings the dumbbells closer.

Initial angular velocity is calculated as 2π radians over 2.0 seconds, equating to π rad/s. Initial moment of inertia measures 3.0 kg•m^2.

Final moment of inertia is 2.2 kg•m^2.

Calculating initial angular momentum:
L = 3.0π

Final angular momentum:
L = 2.2w

Setting initial and final angular momentum equal provides a means to solve for the final angular velocity w:

3.0π = 2.2w
This simplifies to w = 1.4π rad/s.

The expression for rotational energy is:
KE = 0.5Iw^2

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

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

An increase in rotational energy occurs; the origin of this energy is from adjusting the moment of inertia. The professor exerted a radially inward force to pull in the dumbbells, performing work that raises his rotational energy.