Question

Shown below is a small particle of mass 27 g that is moving at a speed of 10 m/s when it collides with and sticks to the edge of a uniform solid cylinder. The cylinder is free to rotate about the axis through its center and perpendicular to the page. The cylinder has a mass of 0.4 kg and a radius of 20 cm and is initially at rest. (a) Find the angular velocity after collision (b) Calculate the final linear velocity of the system

Answer 1

The angular velocity after the collision is 1.93 rad/s, and the final linear velocity of the system is 0.386 m/s.

Step-by-step explanation:

(a) The angular velocity after the collision can be found by applying the principle of conservation of angular momentum. Before the collision, the small particle has an initial momentum of:

Angular momentum = m * v * r = (0.027 kg) * (10 m/s) * (0.2 m) = 0.054 kg⋅m²/s

After the collision, the small particle sticks to the edge of the cylinder, which results in an increase in the moment of inertia. The moment of inertia of the system after the collision can be calculated using the equation:

I = m1 * r1² + I2

where m1, r1, and I2 represent the mass, radius, and moment of inertia of the small particle and the cylinder, respectively.

Substituting the given values into the equation, we get:

I = (0.027 kg) * (0.2 m)² + (0.4 kg) * (0.2 m)² = 0.014 kg⋅m²

Using the principle of conservation of angular momentum:

Angular momentum before = Angular momentum after

m * v * r = I * ω

Substituting the values, we can solve for ω:

(0.027 kg) * (10 m/s) * (0.2 m) = (0.014 kg⋅m²) * ω

ω = (0.027 kg * 10 m/s * 0.2 m) / (0.014 kg⋅m²) = 1.93 rad/s

Therefore, the angular velocity after the collision is 1.93 rad/s.

(b) The final linear velocity of the system can be calculated using the equation:
v = ω * r

Substituting the known values, we get:

v = (1.93 rad/s) * (0.2 m) = 0.386 m/s

Therefore, the final linear velocity of the system is 0.386 m/s.

Answer 2

a) Angular Velocity After Collision:

The angular velocity of the cylinder and the particle after the collision is approximately
`\( 5.95 \, \text{rad/s} \).`

b) Final Linear Velocity of the System:

The final linear velocity of the edge of the cylinder (and the particle attached to it) is approximately
`\( 1.19 \, \text{m/s} \).`

To solve this problem, we can use the principle of conservation of angular momentum. Since there are no external torques acting on the system, the total angular momentum before the collision is equal to the total angular momentum after the collision.

a) Finding the Angular Velocity After Collision

1. Calculate Initial Angular Momentum:

- The particle has linear momentum before the collision, which can be converted to angular momentum about the axis of the cylinder.

- The angular momentum of the particle is
`\( L = m_p * v * r \), where \( m_p = 27 \, \text{g} = 0.027 \, \text{kg} \) (mass of the particle), \( v = 10 \, \text{m/s} \) (velocity of the particle), and \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)`(radius of the cylinder).

2. Initial Angular Momentum of the Cylinder:

- The cylinder is initially at rest, so its angular momentum is zero.

3. Total Initial Angular Momentum:

-
`\( L_{\text{initial}} = L_{\text{particle}} + L_{\text{cylinder}} = m_p * v * r \).`

4. Moment of Inertia of the Cylinder:

The moment of inertia of a solid cylinder about its central axis is
`\( I = (1)/(2) m_c r^2 \), where \( m_c = 0.4 \, \text{kg} \)` is the mass of the cylinder.

5. Final Angular Momentum:

- After the collision, the particle and the cylinder rotate together. The final angular momentum is
`\( L_{\text{final}} = (I + m_p r^2) \omega \), where \( \omega \)`is the angular velocity after the collision.

6. Conservation of Angular Momentum:

`\( L_{\text{initial}} = L_{\text{final}} \)`

`\( m_p * v * r = (I + m_p r^2) \omega \)`

7. Solve for
`\( \omega \)`:

`\( \omega = (m_p * v * r)/(I + m_p r^2) \)`

b) Calculating the Final Linear Velocity of the System

1. Final Angular Velocity:

- Use
`\( \omega \)` found in the previous steps.

2. Final Linear Velocity:

- The linear velocity of the edge of the cylinder (and the particle stuck to it) is
`\( v_{\text{final}} = \omega * r \).`