6. Two disks are rotating about the same axis. Disk A has a moment of inertia of 4.00 kg.m² and an angular velocity of +6.80 rad/s. Disk B is rotating with an angular velocity of -11.8 rad/s. The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.50 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B? kg.m2 36 skp

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### Physics Problem: Rotational Motion of Disks

**Problem Statement:**

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(4.00 \, \text{kg.m}^2\) and an angular velocity of \(+6.80 \, \text{rad/s}\). Disk B is rotating with an angular velocity of \(-11.8 \, \text{rad/s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.50 \, \text{rad/s}\). The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

**Calculation:**

To find the moment of inertia of disk B, we can use the law of conservation of angular momentum, which states that the total angular momentum before the disks are combined is equal to the total angular momentum after they are combined, because no external torques are acting on the system.

The total angular momentum \( L \) is given by:
\[ L = I \cdot \omega \]

Where:
- \( I \) is the moment of inertia
- \( \omega \) is the angular velocity

Initially:
\[ L_{\text{initial}} = I_A \cdot \omega_A + I_B \cdot \omega_B \]

After combining:
\[ L_{\text{final}} = (I_A + I_B) \cdot \omega_f \]

Equating the initial and final angular momentum:
\[ I_A \cdot \omega_A + I_B \cdot \omega_B = (I_A + I_B) \cdot \omega_f \]

Given:
- \( I_A = 4.00 \, \text{kg.m}^2 \)
- \( \omega_A = +6.80 \, \text{rad/s} \)
- \( \omega_B = -11.8 \, \text{rad/s} \)
- \( \omega_f = -2.50 \, \text{rad/s} \)

Substitute these values into the equation:
\[ (4.00 \, \text{kg.m}^2 \cdot 6.80 \, \text{rad/s}) + (I_B \cdot -11.8 \,
Transcribed Image Text:### Physics Problem: Rotational Motion of Disks **Problem Statement:** Two disks are rotating about the same axis. Disk A has a moment of inertia of \(4.00 \, \text{kg.m}^2\) and an angular velocity of \(+6.80 \, \text{rad/s}\). Disk B is rotating with an angular velocity of \(-11.8 \, \text{rad/s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.50 \, \text{rad/s}\). The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B? **Calculation:** To find the moment of inertia of disk B, we can use the law of conservation of angular momentum, which states that the total angular momentum before the disks are combined is equal to the total angular momentum after they are combined, because no external torques are acting on the system. The total angular momentum \( L \) is given by: \[ L = I \cdot \omega \] Where: - \( I \) is the moment of inertia - \( \omega \) is the angular velocity Initially: \[ L_{\text{initial}} = I_A \cdot \omega_A + I_B \cdot \omega_B \] After combining: \[ L_{\text{final}} = (I_A + I_B) \cdot \omega_f \] Equating the initial and final angular momentum: \[ I_A \cdot \omega_A + I_B \cdot \omega_B = (I_A + I_B) \cdot \omega_f \] Given: - \( I_A = 4.00 \, \text{kg.m}^2 \) - \( \omega_A = +6.80 \, \text{rad/s} \) - \( \omega_B = -11.8 \, \text{rad/s} \) - \( \omega_f = -2.50 \, \text{rad/s} \) Substitute these values into the equation: \[ (4.00 \, \text{kg.m}^2 \cdot 6.80 \, \text{rad/s}) + (I_B \cdot -11.8 \,
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