1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the course. If the hoop has a radius of 75.0 cm, what is the moment of inertia of Alex's rolling hoop?

a) 1,970 kg ⋅ m2
b) 26.3 kg ⋅ m2
c) 0.263 kg ⋅ m2
d) 0.197 kg ⋅ m2

2. Jessica stretches her arms out 0.8 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins around on an ice rink at 1.2 m/s.
a. What is the combined angular momentum of the masses?
b. If she pulls her arms in to 0.12 m, what is her new linear speed?

Respuesta :

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 [tex]kg \cdot cm^2[/tex]

Explanation:

Given:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

To Find:

The moment of inertia of Alex's rolling hoop = ?

Solution:

The moment of inertia  = [tex]mr^2[/tex]

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia  = [tex](0.350)(0.75)^2[/tex]

=> moment of inertia  = [tex](0.350)(0.5625)[/tex]

=> moment of inertia  = [tex](0.197)[/tex]

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 [tex]kg m^2 s^{-1}[/tex]

If she pulls her arms in to 0.12 m, her new linear speed  is  [tex]18.33 m/s^2[/tex]

Explanation:

Given:

Mass  = 2.0 kg

Radius = 0.8 m

Velocity =  1.2 m/s

a.The combined angular momentum of the masses:

[tex]L = r \cdot m \cdot v_1[/tex]

Substituting the values,

[tex]L = 0.8 \cdot 2.0 \cdot 1.1[/tex]

L= 1.76 [tex]kg m^2 s^{-1}[/tex]

b. If she pulls her arms in to 0.12 m, what is her new linear speed

[tex]0.12 \cdot 0.8 \cdot v_2 = 1.76[/tex]

[tex] 0.096 cdot v_2 = 1.76[/tex]

[tex] v_2 = \frac{1.76}{0.096}[/tex]

[tex] v_2 = 18.33 m/s^2[/tex]

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