Problem 71 - Fundamentals of Physics

Question

A constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(12 \mathrm{~N}\) is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is \(10 \mathrm{~kg},\) its radius is \(0.10 \mathrm{~m}\) and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?

Step-by-Step Solution

Answer

a = 0.5 m/s², \( \alpha = 5 \text{ rad/s}^2 \), \( \vec{f} = 5 \text{ N} \).

Step 1: Understanding the Problem

We are given a cylinder of mass 10 kg with a radius of 0.10 m, on which a 12 N force is applied horizontally via a fishing line wrapped around it. The cylinder rolls without slipping on a surface, meaning there's both translational and rotational motion involved. We need to find the linear and angular accelerations and the friction force.

Step 2: Identify the Equations

Apply Newton's Second Law for translation: \[ F_{ ext{net}} = m imes a \ F_{ ext{app}} - f = m imes a \] where:- \( F_{\text{net}} \) is the net force,- \( F_{\text{app}} \) is the applied force,- \( f \) is the friction force,- \( a \) is the linear acceleration of the center of mass. For rotation, use:\[\tau = I \times \alpha \ f imes r = \frac{1}{2} m r^2 \alpha \]where:- \( \tau \) is the torque,- \( I \) is the moment of inertia of the cylinder,- \( r \) is the cylinder's radius,- \( \alpha \) is the angular acceleration.

Step 3: Solve for Acceleration (a)

Start with the rotational aspect knowing the cylinder rolls without slipping:\[ a = r \times \alpha \]Plug it into the second equation:\[ f \cdot 0.1 = 0.5 \times 10 \times (0.1)^2 \times \alpha \]Solving gives:\[ \alpha = \frac{20f}{0.1} \quad \text{and hence} \quad a = 0.1 \times \alpha \]Using translation equation:\[ 12 - f = 10 \times a \]Combining gives:\[ 12 - \frac{a}{0.1} \times 0.5 = 10 \times a \ a = 0.5 \frac{m}{s^2}\]

Step 4: Solve for Angular Acceleration (b)

From the rolling condition \(a = r \times \alpha\):\[ \alpha = \frac{a}{r} = \frac{0.5}{0.1} = 5 \, \text{rad/s}^2 \] Thus, the angular acceleration \( \alpha \) is 5 rad/s².

Step 5: Calculate Friction Force (c)

Using \( f = m \times a \), where \( a = 0.5 \frac{m}{s^2} \):\[ f = 10 \times 0.5 = 5 \text{ N} \]The magnitude of the frictional force is 5 N and it acts opposite to the applied force to prevent slipping.

Key Concepts

Newton's Second Law: Understanding Rolling Motion

Newton's Second Law is fundamental in analyzing rolling motion, as it helps us connect force, mass, and acceleration. When a force is applied to an object, it causes the object to accelerate in the direction of the applied force. In the case of a cylinder rolling on a surface, we must consider both translational and rotational motion.

For translational motion, Newton's Second Law is expressed as:

• \[ F_{\text{net}} = m \times a \]

where \( F_{\text{net}} \) is the net force acting on the cylinder, \( m \) is its mass, and \( a \) is the linear acceleration of its center of mass.

• In this scenario, the net force includes the applied force, \( F_{\text{app}} \), and the friction force, \( f \).

The equation becomes:

• \[ F_{\text{app}} - f = m \times a \]

This relationship helps us determine how fast the object moves across the surface. The key here is to understand that the force applied and the friction both play roles: friction acts in the opposite direction but it's essential to enable rolling without slipping. This complex dance of forces results in rolling motion, which simultaneously involves linear and rotational aspects.

By applying Newton's Second Law, we deduce the linear acceleration of the cylinder, a pivotal step in unpacking the mechanics of rolling objects.

Torque and Angular Acceleration: The Rotational Aspect

To understand rolling motion, we need to delve into the concept of torque and angular acceleration. Torque (\( \tau \)) is essentially a rotational force. It is the force that causes an object to rotate about an axis. Just like force leads to linear acceleration, torque leads to angular acceleration.

For the cylinder, which rolls without slipping, the equation for torque is:

• \[ \tau = I \times \alpha \]

where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

• For a uniform solid cylinder, \( I = \frac{1}{2} m r^2 \).
• The frictional force acts at a distance of the radius \( r \) from the center, so torque can also be expressed as friction times radius:
  \[ \tau = f \times r \].

Substituting this into the torque equation gives us:

• \[ f \times r = \frac{1}{2} m r^2 \alpha \]

Here, the angular acceleration \( \alpha \) can be related to the linear acceleration \( a \) by the relation:

• \[ a = r \times \alpha \]

This means that the cylinder’s rotation (angular acceleration) is directly tied to its rolling (linear acceleration). By using these equations, we can find the angular acceleration of the cylinder, which describes how quickly it spins as it rolls across the surface.

Friction Force: Enabling Rolling Without Slipping

Friction plays a subtle yet critical role in rolling motion, especially when we insist on rolling without slipping. It might seem counterintuitive, but without friction, the cylinder would not roll properly; it would just slide like skidding on ice.

• Friction force (\( f \)) acts at the point of contact between the cylinder and the surface.
• It provides the necessary torque to make the cylinder rotate."

In our scenario, we calculate friction force using the known mass and linear acceleration of the cylinder:

• \[ f = m \times a \]

This frictional force is crucial as it ensures the ideal rolling motion by preventing slippage. While it acts opposite to the applied force's direction, its overall effect is to stabilize the motion. Understanding friction force in rolling scenarios is essential, as it provides the grip that keeps rolling objects on track.

In the physical world, friction manages to balance the forces and instigate rolling, ensuring the smooth transition between linear and rotational motion. Without it, neither the pleasing roll nor the control over the object would be possible.