Question

A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 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

Angular acceleration is \(34.78\, \text{rad/s}^2\), cord acceleration \(8.695\, \text{m/s}^2\), axle force is \(40.0 \, \text{N}\) horizontally. Answers change if pull is upward due to additional vertical force.

Step-by-step solution

Identify Known Values and Variables

The radius of the wheel, \( r = 0.250 \) m, the mass of the wheel \( m = 9.20 \) kg, the force applied \( F = 40.0 \) N. We need to find angular acceleration \( \alpha \), linear acceleration \( a \), and the force exerted by the axle.

Calculate Moment of Inertia

For a solid wheel (disk), the moment of inertia \( I \) about its center is given by the formula: \[ I = \frac{1}{2} m r^2 \]Substituting the values, we find:\[ I = \frac{1}{2} \times 9.20 \times (0.250)^2 = 0.2875 \, \text{kg} \, \text{m}^2 \]

Calculate Angular Acceleration

Using Newton's second law for rotation, torque \( \tau \) is related to angular acceleration \( \alpha \) by:\[ \tau = I \alpha \]The applied force creates a torque \( \tau = F \times r \):\[ \tau = 40.0 \times 0.250 = 10.0 \, \text{Nm} \]Solving for \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{10.0}{0.2875} \approx 34.78 \, \text{rad/s}^2 \]

Calculate Linear Acceleration of the Cord

The linear acceleration \( a \) of the pulled cord is related to the angular acceleration by the relationship:\[ a = r \alpha \]Substituting for \( r \) and \( \alpha \):\[ a = 0.250 \times 34.78 = 8.695 \, \text{m/s}^2 \]

Calculate Force Exerted by the Axle

Since there are no horizontal forces exerted by the axle other than balancing the applied force (because the system is frictionless and only one force acts externally), the horizontal force exerted by the axle \( F_{axle} \) is:\[ F_{axle} = F = 40.0 \, \text{N} \]The direction is opposite to the applied force to maintain equilibrium. Also, no vertical forces are exerted because the pull is horizontal.

Analyze Change in Force if Pull is Upward

If the pull were upward, the horizontal components of dynamics don't change, but we introduce a vertical component to balance the gravitational pull and the upward force:- The angular acceleration \( \alpha \) of the wheel and the linear acceleration \( a \) of the cord do not change as these depend solely on the tangential component.- The force exerted by the axle now has a vertical component equal to the downward gravitational force \( mg \).

Key concepts

Angular acceleration

Angular acceleration is the rate at which the angular velocity of an object changes with time. It's the rotational counterpart to linear acceleration. In the context of this exercise, we're dealing with angular acceleration caused by a torque from a steady horizontal pull on the cord wrapped around a wheel. To find the angular acceleration \(\alpha\), we use Newton's second law for rotation, expressed as:

• \( \tau = I \alpha \) where \(\tau\) is the torque, and \(I\) is the moment of inertia.
• The torque \(\tau\) is given by \( F \times r \), where \(F\) is the force applied and \(r\) is the radius.

Substituting the known values into these equations, you would calculate the angular acceleration. Angular acceleration is significant because it quantifies how quickly the object starts spinning faster or slows down its spinning motion.
Understanding angular acceleration helps in various real-world applications, such as designing motors, machinery, and even understanding planetary movements in astrophysics.

Torque

Torque is a measure of the rotational force applied to an object. Think of it as the twist that causes things to spin. The importance of torque lies in its direct relationship to changing an object's rotational motion.
For the wheel described in the exercise,

• The torque \(\tau\) results from the force applied to the cord and is calculated with \( \tau = F \times r \).
• Here, \(F\) represents the magnitude of the force, and \(r\) is the perpendicular distance from the axis of rotation to where the force is applied.

Essentially, torque determines how effectively a force can rotate an object around an axis.
Just like linear force affects linear motion, torque affects rotational motion. This concept is foundational in fields like mechanical engineering and physics, where understanding rotation is crucial to analyzing systems and designing machines.

Newton's second law

Newton's Second Law forms the backbone of understanding both linear and rotational dynamics. Its rotational form is what we're using here to connect torque and angular motion.

• The linear version of Newton's Second Law, \( F = ma \), describes how the force on an object relates to its mass and linear acceleration.
• Analogously, for rotational motion, it morphs to \( \tau = I \alpha \), showing how torque relates to moment of inertia \(I\) and angular acceleration \(\alpha\).

With this law, we can understand how different forces result in changes in motion. For instance, in a frictionless environment as described in the exercise, we see how an applied force causes not only a linear motion of a rope but also results in both angular motion (the wheel spinning) and associated reactions (axial forces).
By bridging the concepts of torque and angular acceleration, Newton's Second Law brings a coherent understanding of how rotational systems behave and predict the dynamics of these systems efficiently.