Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller flywheel of radius 21 cm has a cord that has a pulling force of 50 N on it. What pulling force (in N) needs to be applied to the cord connecting the larger flywheel of radius 34 cm such that the combination does not rotate? E = ? 50 N

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### Problem Description

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis. The smaller flywheel has a radius of 21 cm with a cord exerting a pulling force of 50 N on it. Determine the pulling force (in Newtons) required on the cord connected to the larger flywheel, which has a radius of 34 cm, to ensure the system remains stationary.

### Diagram Explanation

- **Diagram Elements**:
  - There are two concentric circles representing the flywheels, with the smaller flywheel situated inside the larger one.
  - The smaller flywheel (radius 21 cm) is shown on the left, with a horizontal arrow pointing to the right labeled "50 N", indicating the direction and magnitude of the pulling force on this wheel.
  - The larger flywheel (radius 34 cm) is on the right side, with a horizontal arrow pointing to the left labeled "F = ?", representing the unknown force needed to keep the system from rotating.

### Calculations

To solve for \( F \), the force needed on the larger flywheel, the condition for static equilibrium requires that the torque exerted on the smaller flywheel equals the torque exerted on the larger flywheel.

The torque (\( \tau \)) is calculated as the product of the force and the radius:

- Torque on smaller flywheel: \( \tau_1 = 50 \, \text{N} \times 21 \, \text{cm} \)
- Torque on larger flywheel: \( \tau_2 = F \times 34 \, \text{cm} \)

For the system to be in equilibrium, these torques must be equal:

\[ 50 \, \text{N} \times 21 \, \text{cm} = F \times 34 \, \text{cm} \]

Solve for \( F \):

\[ F = \frac{50 \times 21}{34} \]

### Answer

Insert the calculated force \( F \) to maintain equilibrium: \(\boxed{30.88 \, \text{N}}\), rounded to two decimal places.

This pulling force ensures the flywheel system remains stationary, with no net rotation.
Transcribed Image Text:### Problem Description Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis. The smaller flywheel has a radius of 21 cm with a cord exerting a pulling force of 50 N on it. Determine the pulling force (in Newtons) required on the cord connected to the larger flywheel, which has a radius of 34 cm, to ensure the system remains stationary. ### Diagram Explanation - **Diagram Elements**: - There are two concentric circles representing the flywheels, with the smaller flywheel situated inside the larger one. - The smaller flywheel (radius 21 cm) is shown on the left, with a horizontal arrow pointing to the right labeled "50 N", indicating the direction and magnitude of the pulling force on this wheel. - The larger flywheel (radius 34 cm) is on the right side, with a horizontal arrow pointing to the left labeled "F = ?", representing the unknown force needed to keep the system from rotating. ### Calculations To solve for \( F \), the force needed on the larger flywheel, the condition for static equilibrium requires that the torque exerted on the smaller flywheel equals the torque exerted on the larger flywheel. The torque (\( \tau \)) is calculated as the product of the force and the radius: - Torque on smaller flywheel: \( \tau_1 = 50 \, \text{N} \times 21 \, \text{cm} \) - Torque on larger flywheel: \( \tau_2 = F \times 34 \, \text{cm} \) For the system to be in equilibrium, these torques must be equal: \[ 50 \, \text{N} \times 21 \, \text{cm} = F \times 34 \, \text{cm} \] Solve for \( F \): \[ F = \frac{50 \times 21}{34} \] ### Answer Insert the calculated force \( F \) to maintain equilibrium: \(\boxed{30.88 \, \text{N}}\), rounded to two decimal places. This pulling force ensures the flywheel system remains stationary, with no net rotation.
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