Question

In the figure here, three particles of mass m = 0.022 kg are fastened to three rods of length d = 0.15 m and negligible mass. The rigid assembly rotates about point O at angular speed ? = 0.50 rad/s. About O, what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the magnitude of the angular momentum of the assembly?

Answer 1

`(a) The rotational inertia of the assembly about point O is \(I = (7)/(12) m d^2\).`

`(b) The magnitude of the angular momentum of the middle particle is \(L_m = (1)/(4) m d^2 \omega\) where \(\omega\) is the angular speed.`

`(c) The magnitude of the angular momentum of the assembly about point O is \(L_{\text{assembly}} = (7)/(12) m d^2 \omega\).`

Step-by-step explanation:

In rotational dynamics the rotational inertia I of a system is analogous to mass in linear motion. For a system of particles rotating about a fixed axis the rotational inertia I is given by
`\(I = \sum m_i r_i^2\)` where m_i is the mass of each particle and r_i is the perpendicular distance from the axis. In this case with three particles of mass m attached to rods of length d and rotating about point O the rotational inertia I is calculated as
`\(I = (7)/(12) m d^2\)`.

The angular momentum L of a particle about a fixed axis is given by L = I omega where omega is the angular speed. For the middle particle in this assembly the angular momentum
`(\(L_m\)) is \(L_m = (1)/(4) m d^2 \omega\)`. This accounts for the particular arrangement and distribution of mass in the assembly.

The angular momentum of the entire assembly about point O
`(\(L_{\text{assembly}}\))` is the sum of the angular momenta of its individual particles. For the given assembly L_text assembly = frac7 12 m d^2 omega. This result is derived from the collective contribution of the three particles and their respective distances from the axis of rotation emphasizing the importance of understanding the distribution of mass in rotational systems.

Answer 2

(a) The rotational inertia of the assembly is
`\( 0.00151875 \, \text{kg} \cdot \text{m}^2 \).`

(b) The magnitude of the angular momentum of the middle particle is
`\( 0.000253125 \, \text{kg} \cdot \text{m}^2/\text{s} \).`

(c) The magnitude of the angular momentum of the assembly is
`\( 0.000759375 \, \text{kg} \cdot \text{m}^2/\text{s} \).`

(a) The rotational inertia of the assembly:

The formula for the rotational inertia of point masses rotating about an axis is
`\( I = \sum m_i r_i^2 \), where \( m_i \)` is the mass of each point particle and
`\( r_i \)` is its distance from the axis of rotation.

Given that the mass of each particle is
`\( m = 0.022 \, \text{kg} \)` and the length of each rod is
`\( d = 0.15 \, \text{m} \)`, the distance of each particle from the axis of rotation is
`\( (d)/(2) \).`

The rotational inertia for each particle about point O is
`\( I_i = m_i \cdot \left((d)/(2)\right)^2 = m \cdot \left((d)/(2)\right)^2 \).`

Since there are three particles, the total rotational inertia of the assembly is
`I_(total) = 3* i_i`

Plug in the values to find
`\( I_{\text{total}} \)`:

`\( I_i = 0.022 \, \text{kg} * \left(\frac{0.15 \, \text{m}}{2}\right)^2 \)`

`\( I_i = 0.00050625 \, \text{kg} \cdot \text{m}^2 \)`

`\( I_{\text{total}} = 3 * 0.00050625 \, \text{kg} \cdot \text{m}^2 \)`

`\( I_{\text{total}} = 0.00151875 \, \text{kg} \cdot \text{m}^2 \)`

(b) The magnitude of the angular momentum of the middle particle:

The formula for angular momentum is
`\( L = I \cdot \omega \)`, where I is the rotational inertia and
`\( \omega \)` is the angular speed.

For the middle particle, the rotational inertia is \( I_i = m \cdot \left(\frac{d}{2}\right)^2 \) as calculated before.

Angular momentum of the middle particle
`\( L_{\text{middle}} = I_i \cdot \omega \).`

`\( L_{\text{middle}} = 0.00050625 \, \text{kg} \cdot \text{m}^2 * 0.50 \, \text{rad/s} \)`

`\( L_{\text{middle}} = 0.000253125 \, \text{kg} \cdot \text{m}^2/\text{s} \)`

(c) The magnitude of the angular momentum of the assembly:

Since the three particles rotate together, the angular momentum of the assembly is the sum of the angular momenta of all three particles.

`\( L_{\text{assembly}} = 3 * L_{\text{middle}} \)`

`\( L_{\text{assembly}} = 3 * 0.000253125 \, \text{kg} \cdot \text{m}^2/\text{s} \)`

`\( L_{\text{assembly}} = 0.000759375 \, \text{kg} \cdot \text{m}^2/\text{s} \)`