Question

A 5.00-kg solid sphere with a radius of 3.00 m rotates about a line passing through its center at a rate of 10.0 rad/s. What is the component of the sphere's angular momentum measured in kg m²/s about the same line?

Answer 1

The component of the sphere's angular momentum about the line passing through its center is 180 kg m²/s, calculated using the formula L = Iω with a moment of inertia for a solid sphere I = (2/5)mr².

Step-by-step explanation:

To calculate the component of the sphere's angular momentum about the axis of rotation, you can use the formula for the angular momentum of a rotating object, which is L = Iω. Here, I represents the moment of inertia of the object, and ω represents the angular velocity. For a solid sphere, the moment of inertia is given by I = (2/5)mr², where m is the mass of the sphere and r is the radius. With a mass of 5.00 kg and a radius of 3.00 m, and rotating at 10.0 rad/s, the angular momentum L can be found as follows:

I = (2/5)(5 kg)(3 m)² = (2/5)(5 kg)(9 m²) = 18 kg m²

L = Iω = 18 kg m² × 10.0 rad/s = 180 kg m²/s

Therefore, the component of the sphere's angular momentum measured in kg m²/s about the line passing through its center is 180 kg m²/s.

Answer 2

The component of the sphere's angular momentum is
`\(900.0 \, \text{kg} \, \text{m}^2/\text{s}\) about the same line.`

The angular momentum
`(\(L\))` of a rotating object can be calculated using the formula:

`\[L = I \cdot \omega\]`

Where:

-
`\(L\)` is the angular momentum (in kg m²/s),

-
`\(I\)` is the moment of inertia of the object,

-
`\(\omega\)` is the angular velocity (in rad/s).

The moment of inertia
`(\(I\))` for a solid sphere rotating about an axis passing through its center is given by:

`\[I = (2)/(5) m r^2\]`

Where:

-
`\(m\)` is the mass of the sphere (in kg),

-
`\(r\)` is the radius of the sphere (in meters).

Given:

- Mass of the sphere
`(\(m\)`) = 5.00 kg

- Radius of the sphere
`(\(r\))` = 3.00 m

- Angular velocity
`(\(\omega\))` = 10.0 rad/s

Let's calculate the moment of inertia
`(\(I\))` first. Then, we can find the angular momentum
`(\(L\))` component.

First, calculate the moment of inertia
`(\(I\))` using the formula:

`\[I = (2)/(5) m r^2\]`

Substitute the given values:

`\[I = (2)/(5) * 5.00 \, \text{kg} * (3.00 \, \text{m})^2\]`

Now, calculate
`\(I\).`

Let's calculate the moment of inertia
`(\(I\)):`

`\[I = (2)/(5) * 5.00 \, \text{kg} * (3.00 \, \text{m})^2\]`

Now, calculate
`\(I\).`

The moment of inertia
`(\(I\))` for the solid sphere is:

`\[I = (2)/(5) * 5.00 \, \text{kg} * (3.00 \, \text{m})^2 = 90.00 \, \text{kg} \, \text{m}^2\]`

Now, we can calculate the angular momentum
`(\(L\))` component using the formula:

`\[L = I \cdot \omega\]`

Substitute the given angular velocity
`(\(\omega\)):`

`\[L = 90.00 \, \text{kg} \, \text{m}^2 \cdot 10.0 \, \text{rad/s}\]`

Now, calculate
`\(L\).`

The angular momentum
`(\(L\))` component of the solid sphere about the given axis is:

`\[L = 90.00 \, \text{kg} \, \text{m}^2 \cdot 10.0 \, \text{rad/s} = 900.0 \, \text{kg} \, \text{m}^2/\text{s}\]`

So, the component of the sphere's angular momentum is
`\(900.0 \, \text{kg} \, \text{m}^2/\text{s}\) about the same line.`