Question

An overhead view of a 2.60 kg plastic rod of length 1.20 m on a table. one end of the rod is attached to the table, and the rod is free to pivot about this point without friction. a disk of mass 47.0 g slides toward the opposite end of the rod with an initial velocity of 32.5 m/s. the disk strikes the rod and sticks to it. after the collision, the rod rotates about the pivot point. a thin vertical rod has a point labeled pivot at its bottom, and a disk moving horizontally along an arrow labeled vector v that will reach the rod perpendicularly at the top. (a) what is the angular velocity, in rad/s, of the two after the collision?

Answer 1

After the collision, the angular velocity (ω) of the rod and the disk combined is approximately 1.39 rad/s.

How to find angular velocity?

Calculate the initial angular momentum (
`\( L_{\text{initial}} \)`):

The linear momentum of the disk before the collision is

`\( p = m_{\text{disk}} * v \)`.

`\( m_{\text{disk}} = 47.0 \, \text{g} \\= 0.047 \, \text{kg} \)` (conversion from grams to kilograms).

`\( v = 32.5 \, \text{m/s} \)` (initial velocity of the disk).

The distance from the pivot to the point of impact is equal to the length of the rod,

`\( r = L = 1.20 \, \text{m} \)`.

Therefore,

`\( L_{\text{initial}} = L * m_{\text{disk}} * v \\= 1.20 * 0.047 * 32.5 \)`.

Calculate the moment of Inertia (I) of the system after the collision:

Moment of inertia of the rod

(
`\( I_{\text{rod}} \)`) =
`\( (1)/(3) m_{\text{rod}} L^2 \)`

`\( m_{\text{rod}} = 2.60 \, \text{kg} \)`

Therefore,

`\( I_{\text{rod}} = (1)/(3) * 2.60 * 1.20^2 \).`

Moment of inertia of the disk

`(\( I_{\text{disk}} \)) = \( m_{\text{disk}} * L^2 \)`.

Therefore,

`\( I_{\text{disk}} = 0.047 * 1.20^2 \).`

Total moment of inertia
`(\( I_{\text{total}} \)) = \( I_{\text{rod}} + I_{\text{disk}} \)`

Using conservation of angular momentum to find the final angular velocity (
`\( \omega \)`):

`\( L_{\text{initial}} = I_{\text{total}} * \omega \).`

Solve for ω:

`\( \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} \)`.

Initial angular momentum (
`\( L_{\text{initial}} \)`):

`\[ L_{\text{initial}} = 1.20 * 0.047 * 32.5 \\= 1.833 \, \text{kg}\cdot\text{m}^2/\text{s} \]`

Moment of inertia for rod (
`\( I_{\text{rod}} \)`):**

`\[ I_{\text{rod}} = (1)/(3) * 2.60 * 1.20^2 \\= 1.248 \, \text{kg}\cdot\text{m}^2 \]`

Moment of inertia for disk (
`\( I_{\text{disk}} \)`):**

`\[ I_{\text{disk}} = 0.047 * 1.20^2 \\= 0.06768 \, \text{kg}\cdot\text{m}^2 \]`

Total moment of inertia (
`\( I_{\text{total}} \)`):**

`\[ I_{\text{total}} = I_{\text{rod}} + I_{\text{disk}} \\ =1.31568 \, \text{kg}\cdot\text{m}^2 \]`

Final angular velocity (ω):

`\[ \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} \\ = (1.833)/(1.31568) \\ = 1.39 \, \text{rad/s} \]`

Therefore, after the collision, the angular velocity of the rod-disk system is approximately 1.39 rad/s.