Question

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?

Answer 1

Question 1: The moment of inertia of Alex's rolling hoop is 0.263 kg ⋅ m^2. Question 2: The combined angular momentum of the masses is 1.92 kg ⋅ m^2/s. The new linear speed after pulling her arms in is 0.90 m/s.

Step-by-step explanation:

Question 1:

The moment of inertia of Alex's rolling hoop can be calculated using the formula:

I = 0.5 * m * r^2,

where I is the moment of inertia, m is the mass of the hoop, and r is the radius of the hoop.

Plugging in the values, we have:

I = 0.5 * 0.350 kg * (0.75 m)^2 = 0.263 kg ⋅ m^2.

Therefore, the moment of inertia of Alex's rolling hoop is 0.263 kg ⋅ m^2 (option c).

Question 2:

a) The combined angular momentum of the masses can be calculated using the formula:

L = m * r * v,

where L is the angular momentum, m is the mass, r is the distance from the axis of rotation, and v is the linear speed.

Plugging in the values, we have:

L = 2.0 kg * 0.800 m * 1.2 m/s = 1.92 kg ⋅ m^2/s.

Therefore, the combined angular momentum of the masses is 1.92 kg ⋅ m^2/s.

b) The new linear speed can be calculated using the formula:

v' = rω,

where v' is the new linear speed, r is the new radius, and ω is the new angular velocity.

Plugging in the values, we have:

v' = 0.12 m * (2π rad/rev) * (1.2 rev/s) = 0.90 m/s.

Therefore, the new linear speed is 0.90 m/s.

Answer 2

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197
`kg \cdot cm^2`

Step-by-step 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 =
`mr^2`

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 =
`(0.350)(0.75)^2`

=> moment of inertia =
`(0.350)(0.5625)`

=> moment of inertia =
`(0.197)`

Question 2:

Answer:

The combined angular momentum of the masses is 1.76
`kg m^2 s^(-1)`

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

Step-by-step explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

`L = r \cdot m \cdot v_1`

Substituting the values,

`L = 0.8 \cdot 2.0 \cdot 1.1`

L= 1.76
`kg m^2 s^(-1)`

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

`0.12 \cdot 0.8 \cdot v_2 = 1.76`

`0.096 cdot v_2 = 1.76`

`v_2 = (1.76)/(0.096)`

`v_2 = 18.33 m/s^2`