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Kipish [7]
3 years ago
7

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.A) Estimate the skater's moment of inertia to two significant figures.B)If she were to hold her arms outward, rather than at her sides, would her moment of inertia increase, decrease, or remain unchanged?
Physics
1 answer:
Oxana [17]3 years ago
6 0

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

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Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

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The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

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