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Chapter 10: Problem 12

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?

Short Answer

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

Step by step solution

01

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:\[ I = \frac{1}{2}(9.20 \, \text{kg})(0.250 \, \text{m})^2 \approx 0.2875 \, \text{kg} \cdot \text{m}^2 \]
02

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 \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{10.0 \, \text{Nm}}{0.2875 \, \text{kg} \cdot \text{m}^2} \approx 34.78 \, \text{rad/s}^2 \]
03

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:\[ a = 0.250 \, \text{m} \times 34.78 \, \text{rad/s}^2 \approx 8.695 \, \text{m/s}^2 \]
04

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:\[ f = 40.0 \, \text{N} - 80 \, \text{N} = -40 \, \text{N} \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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鈥檚 linked to the wheel鈥檚 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/s虏 from previous calculation),
  • \( r \) is wheel鈥檚 radius (0.250 m).
Subsequently, \( a \) becomes approximately 8.695 m/s虏. 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.

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Most popular questions from this chapter

A horizontal plywood disk with mass 7.00 \(\mathrm{kg}\) and diameter 1.00 \(\mathrm{m}\) pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 m to the disk. A 1.20 -kg, battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 \(\mathrm{m} / \mathrm{s}\) relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

While exploring a castle, Exena the Exterminator is spotted by a dragon that chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10 . (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

In a lab experiment you let a uniform ball roll down a curved track. The ball starts from rest and rolls without slipping.While on the track, the ball descends a vertical distance \(h .\) The lower end of the track is horizontal and extends over the edge of the lab table; the ball leaves the track traveling horizontally. While free-falling after leaving the track, the ball moves a horizontal distance \(x\) and a vertical distance \(y\) . (a) Calculate \(x\) in terms of \(h\) and \(y\) ignoring the work done by friction. (b) Would the answer to part (a) be any different on the moon? (c) Although you do the experiment very carefully, your measured value of \(x\) is consistently a bit smaller than the value calculated in part (a). Why? (d) What would \(x\) be for the same \(h\) and \(y\) as in part (a) if you let a silver dollar roll down the track? You can ignore the work done by friction.
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