Problem 11 - Fundamentals of Physics

Question

A constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(10 \mathrm{~N}\) is applied to a wheel of mass \(10 \mathrm{~kg}\) and radius \(0.30 \mathrm{~m}\). The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude \(0.60 \mathrm{~m} / \mathrm{s}^{2}\). (a) In unit-vector notation, what is the frictional force on the wheel? (b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

Step-by-Step Solution

Answer

(a) \(-4 \, \text{N} \hat{i}\); (b) \(0.6 \, \text{kg} \cdot \text{m}^2\).

Step 1: Identify and Write Down Given Values

We have: force \( \vec{F}_{\text{app}} = 10 \, \text{N} \), mass of the wheel \( m = 10 \, \text{kg} \), radius \( r = 0.3 \, \text{m} \), and the acceleration of the center of mass \( a = 0.6 \, \text{m/s}^2 \). The force of friction is the unknown, and the rotational inertia \( I \) is to be determined.

Step 2: Apply Newton’s Second Law for Translation

According to Newton's Second Law, the net force on the wheel equals the mass times the acceleration of the center of mass: \( F_{\text{net}} = m \cdot a \). Write the equation: \( \vec{F}_{\text{app}} - \vec{F}_{\text{friction}} = m \cdot a \). Substitute known values: \( 10 \, \text{N} - \vec{F}_{\text{friction}} = 10 \, \text{kg} \times 0.6 \, \text{m/s}^2 \). Solve for \( \vec{F}_{\text{friction}} \).

Step 3: Solve for Frictional Force

Re-arrange the equation from Step 2: \( \vec{F}_{\text{friction}} = 10 \, \text{N} - 6 \, \text{N} \). Calculate: \( \vec{F}_{\text{friction}} = 4 \, \text{N} \). The frictional force acts opposite to the direction of \( \vec{F}_{\text{app}} \), so in unit-vector notation: \( \vec{F}_{\text{friction}} = -4 \, \text{N} \hat{i} \).

Step 4: Apply Torque Equation to Solve for Rotational Inertia

Torque \( \tau \) is related to rotational inertia \( I \) and angular acceleration \( \alpha \) by the equation \( \tau = I \cdot \alpha \). The torque due to friction is \( \tau = r \cdot F_{\text{friction}} \). Simultaneously, \( a = \alpha \cdot r \), so \( \alpha = \frac{a}{r} \). Substitute \( \tau = I \cdot \frac{a}{r} \) into \( I \cdot \frac{a}{r} = r \cdot F_{\text{friction}} \).

Step 5: Solve for Rotational Inertia

Substitute known values: \( I \cdot \frac{0.6 \, \text{m/s}^2}{0.3 \, \text{m}} = 0.3 \, \text{m} \cdot 4 \, \text{N} \). Simplify and solve for \( I \): \( I \cdot 2 = 1.2 \, \text{N} \cdot \text{m} \), giving \( I = 0.6 \, \text{kg} \cdot \text{m}^2 \).

Step 6: Express the Final Answers

For part (a), the frictional force is \( \vec{F}_{\text{friction}} = -4 \, \text{N} \hat{i} \). For part (b), the rotational inertia about the center of mass is \( I = 0.6 \, \text{kg} \cdot \text{m}^2 \).

Key Concepts

Frictional Force

Frictional force is a crucial concept in physics, especially when discussing objects in motion. It is the force that opposes the movement or the intended movement of an object in contact with a surface. In this exercise, we examined the frictional force acting on a wheel moving smoothly over a surface. Friction is essential here because it provides the necessary grip for the wheel to roll without slipping. Without it, the wheel would not rotate effectively, thus failing to convert the applied force into rotational motion.

• Frictional force can be static or kinetic, depending on whether the object is at rest or in motion.
• In this problem, the static frictional force prevents slipping, allowing smooth rolling.
• According to the given solution, the force of friction opposing the applied force was determined to be 4 N.

Understanding frictional force and its role is vital for solving problems involving rolling motion, as it affects how forces are applied and how they result in motion.

Rotational Inertia

Rotational inertia, sometimes called the moment of inertia, describes an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In simple terms, the farther the mass is from the axis, the higher the rotational inertia. Higher rotational inertia means more torque is required to achieve the same angular acceleration. In the wheel scenario, we calculated the rotational inertia with the formula for torque and known values.

• The formula used is: \[ \tau = I \cdot \alpha \]
• Rotational inertia (I) was found to be 0.6 kg·m².
• This value represents how much the wheel resists being spun by applied forces.

Understanding rotational inertia helps in designing rolling objects like wheels, where efficiency and stability in rotational motion are essential.

Torque

Torque is the rotational equivalent of linear force. It measures how much a force acts on an object to make it rotate, which depends not only on the force's magnitude but also on the lever arm distance from the axis of rotation. This lever arm is often the radius in problems involving wheels. When solving this exercise, torque due to friction was considered. We linked it to angular acceleration to find the rotational inertia.

• Torque ( \( \tau \)) is given by the equation: \[ \tau = r \cdot F \]
• For a wheel, torque depends on frictional force applied at a distance (radius) from the center.
• Understanding torque helps explain how forces cause twisting and turning effects.

Torque plays a pivotal role in mechanical design and analysis, guiding the understanding of how effectively forces applied at a distance cause rotational effects.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. It describes how quickly an object speeds up or slows down its rotation. In conversion from linear to rotational motion, understanding angular acceleration is necessary to relate the linear acceleration of the wheel’s center of mass with its spin.

• In the problem, angular acceleration ( \( \alpha \)) is obtained from \[ a = \alpha \cdot r \]
• It uses the relationship between linear acceleration and radius.
• Helps in finding how the wheel's movement translates into rotation speed changes.

Comprehending angular acceleration is vital for problems dealing with circular motion, as it links various aspects of motion, illustrating how linear forces translate into rotational effects.