Problem 12 - University Physics with Modern Physics

Question

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg} .\) A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

Step-by-Step Solution

Answer

Angular acceleration is 34.78 rad/s², cord acceleration is 8.695 m/s², and axle force is 40 N horizontal. Changing pull direction would alter axle force and torque application.

Step 1: Calculate Moment of Inertia

The wheel is a solid disk with radius \( r = 0.250 \ \text{m} \) and mass \( m = 9.20 \ \text{kg} \). The moment of inertia for a solid disk about its center is given by \( I = \frac{1}{2}mr^2 \). Substitute the given values:\[ I = \frac{1}{2}(9.20 \, \text{kg})(0.250 \, \text{m})^2 \approx 0.2875 \, \text{kg} \cdot \text{m}^2 \]

Step 2: Compute Angular Acceleration

The force exerted is \( F = 40.0 \, \text{N} \), and this force generates a torque about the center of the wheel. Torque \( \tau \) is given by \( \tau = rF \) because the force is tangent to the wheel and has a perpendicular lever arm equal to the radius. Substitute the known values:\[ \tau = 0.250 \, \text{m} \times 40.0 \, \text{N} = 10.0 \, \text{Nm} \]The angular acceleration \( \alpha \) is related to torque by \( \tau = I\alpha \). Solve for \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{10.0 \, \text{Nm}}{0.2875 \, \text{kg} \cdot \text{m}^2} \approx 34.78 \, \text{rad/s}^2 \]

Step 3: Calculate Linear Acceleration of the Cord

The linear acceleration \( a \) of the cord is related to the angular acceleration by \( a = r\alpha \). Substitute the values:\[ a = 0.250 \, \text{m} \times 34.78 \, \text{rad/s}^2 \approx 8.695 \, \text{m/s}^2 \]

Step 4: Determine Force on the Wheel by the Axle

The wheel is subjected to the force along the horizontal direction as well as the reaction force by the axle, supporting it along the horizontal due to the pull. The net horizontal force acting on the wheel comes from the force exerted and the inertia force due to the rotation of the wheel.First, consider Newton's second law: \( F - f = ma \)Here, \( F \) is the exerted force (40.0 N) and \( f \) is the reaction force by the axle, while \( a \) is the linear acceleration we solved in Step 3. Using the linear acceleration gotten:- The acceleration component \( ma = 9.20 \, \text{kg} \times 8.695 \, \text{m/s}^2 \approx 80 \, \text{N} \).Thus, the horizontal force reaction exerted by the axle is in the opposite direction:\[ f = 40.0 \, \text{N} - 80 \, \text{N} = -40 \, \text{N} \]

Step 5: Analyze Effect of Changing Pull Direction

If the pull were upward instead of horizontal, only Step 4 would change. This is because the torque and moment of inertia remain the same, so the angular and linear accelerations (in Steps 2 and 3) do not change. In Step 4, the direction of the reaction force by the axle would be different, needing a vertical component instead of horizontal.

Key Concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational motion, similar to mass in linear motion. It quantifies how difficult it is to change the angular velocity of an object. For a wheel or disk, the moment of inertia depends on both its mass and the square of its radius. It's important because it determines how much torque is needed for a desired angular acceleration.

The formula for the moment of inertia of a solid disk revolving around its center is: \[ I = \frac{1}{2}mr^2 \]Where:

• \( I \) is the moment of inertia,
• \( m \) represents the mass of the wheel (in this case 9.20 kg),
• \( r \) is the radius (here 0.250 m).

For our wheel, the calculation gives us \( I \approx 0.2875 \text{ kg} \cdot \text{m}^2 \). This value tells us how the mass is spread out from the axis and affects the wheel’s rotation.

Torque

Torque is the rotational equivalent of force. It measures the tendency of a force to rotate an object around an axis. In this situation, the torque is generated by the horizontal pull exerted on the cord wrapped around the wheel.

The equation for torque when force is applied tangentially is:\[ \tau = rF \]Here:

• \( \tau \) is the torque,
• \( r \) is the radius of the wheel (0.250 m),
• \( F \) is the pulling force (40.0 N).

Substituting the values, we find \( \tau = 10.0 \text{ Nm} \). Torque is crucial because it affects the angular acceleration of the wheel, highlighting how rotational balancing acts are performed.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. It describes how quickly the wheel gains rotational speed and is directly influenced by the torque and the moment of inertia.

The relationship is given by:\[ \tau = I\alpha \] where \( \alpha \) (alpha) is the angular acceleration. To calculate \( \alpha \):\[ \alpha = \frac{\tau}{I} \]For our wheel, substituting \( \tau = 10.0 \text{ Nm} \) and \( I = 0.2875 \text{ kg} \cdot \text{m}^2 \), we find:\[ \alpha \approx 34.78 \text{ rad/s}^2 \].Angular acceleration is not just a measure for engineers. It tells us how dynamic and responsive a rotating system will be under a given force.

Linear Acceleration

Linear acceleration relates to how quickly the cord being pulled accelerates as it unspools. It’s linked to the wheel’s angular acceleration via the radius.The relation can be expressed as:\[ a = r\alpha \]Where:

• \( a \) is linear acceleration,
• \( \alpha \) is angular acceleration (34.78 rad/s² from previous calculation),
• \( r \) is wheel’s radius (0.250 m).

Subsequently, \( a \) becomes approximately 8.695 m/s². Linear acceleration shows the effect of rotational movement on linear motion, making it a key concept for understanding how systems translate rotation into linear movement.