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sashaice [31]
3 years ago
13

A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outst

retched hands are 0.785 m from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to 32.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in
Physics
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

r = 0.491 m

Explanation:

In this case the System is formed by the teacher with the two masses, so the forces during movement are internal and the angular momentum is conserved

Initial.

         L₀ = I₀ w₀

Final

         Lf = I w

        L₀ = Lf

       I₀ w₀ = I w

The moment of inertia is

      I₀ = m r₀²

      I = m r²

       

Let's replace

       m r₀² w₀ = m r² w

       r²  = r₀² w₀ / w

Angular velocity

       w₀ = 10 rpm (2pi rad / 1 rev) (1 min / 60 s) = 1.047 rad / s

       w = 32.5 rpm = 3.403 rad / s

Let's calculate

      r = √(0.785 1.047 / 3.403)

      r = 0.491 m

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Semmy [17]

Answer:

9.6 rad/s

Explanation:

L = length of the metal rod = 50 cm = 0.50 m

M = Mass of the long metal rod = 780 g = 0.780 kg

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\tau = 250 Nm

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w = Angular velocity after the blow

Using Impulse-change in angular momentum, we have

I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}

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Water flows into a swimming pool at the rate of 5.85 gal/min. If the pool dimensions are 21.2 ft wide, 46.1 ft long and 19.4 ft
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