Question

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

Short answer

The acceleration is \( a = \frac{F}{2M} \) and the friction force is \( f = \frac{F}{2} \).

Step-by-step solution

Establish Equations of Motion

A horizonally applied force, \( F \), contributes to the linear and rotational motion of the cylinder. We begin by using Newton's second law for linear motion: \( F - f = Ma \), where \( f \) is the friction force, and \( a \) is the linear acceleration of the center of mass.

Analyze Rotational Motion

Since the cylinder rolls without slipping, the friction force \( f \) is static and provides torque. For rotational motion, use \( \tau = I\alpha \). The torque \( \tau = fR \) and moment of inertia for a hollow cylinder is \( I = MR^2 \). So, \( fR = MR^2\alpha \).

Relate Angular and Linear Acceleration

Since the cylinder rolls without slipping, the linear acceleration \( a \) and the angular acceleration \( \alpha \) are related: \( a = R\alpha \). Substitute for \( \alpha \) from previous step: \( \alpha = \frac{a}{R} \).

Solve for Friction Force

Substitute \( \alpha = \frac{a}{R} \) into the rotational equation: \( fR = MR^2 \frac{a}{R} \rightarrow f = Ma \).

Substitute Friction Force into Linear Motion Equation

Using the expression for friction force from the previous step, substitute \( f = Ma \) into the equation for linear motion: \( F - Ma = Ma \).

Solve for Acceleration

Rearrange the equation \( F - Ma = Ma \) to find \( a: F = 2Ma \rightarrow a = \frac{F}{2M} \).

Determine Friction Force

Substitute \( a = \frac{F}{2M} \) back into the expression for friction force: \( f = Ma = M \cdot \frac{F}{2M} = \frac{F}{2} \).

Key concepts

Newton's Second Law

Newton's second law is a fundamental principle that describes how forces affect motion. The law is expressed as \( F = ma \), where \( F \) is the net force acting on an object, \( m \) is the mass of the object, and \( a \) is the acceleration. This relationship shows how an increase in force can increase acceleration, assuming mass stays constant.
For the lawn roller problem, Newton's second law helps us set up the equation of motion: \( F - f = Ma \). Here, \( F \) is the applied force, \( f \) is the friction force, and \( a \) is the acceleration of the mass \( M \). By applying Newton's second law in this context, we are able to understand how the forces at play determine the linear acceleration of the roller.
By focusing on the balance of forces acting horizontally, we see how the friction force and applied force combine to result in the system's acceleration. This clarity helps us solve for unknowns like the friction force and the system's acceleration.

Friction Force

Friction force is crucial in a rolling motion scenario, especially when the object rolls without slipping. It acts at the point of contact between surfaces, in this case between the lawn roller and the ground. This force is necessary for the rotation of the roller, acting to prevent slipping and ensure there is traction.
Friction can be either static or kinetic, depending on whether the object is moving relative to the surface. In the case of our lawn roller, the friction is static because the roller rolls without slipping. Here, friction does not do any work because the point of contact is momentarily at rest relative to the ground.
In our problem, the friction force \( f \) provides the torque required for rotational motion. It is calculated by setting it equal to the product of mass and acceleration \( Ma \), which results in \( f = \frac{F}{2} \). This shows how the friction strengthens the motion, providing the necessary force to keep the balance with the applied force.

Rotational Dynamics

Rotational dynamics covers how torque and moment of inertia influence the motion of rotating objects. It is similar to linear dynamics but involves quantities like angular acceleration \( \alpha \), torque \( \tau \), and moment of inertia \( I \).
For the lawn roller, the torque is produced by the friction force \( f \) and is calculated as \( \tau = fR \), where \( R \) is the radius of the roller. This torque leads to angular acceleration, expressed with the equation \( \tau = I\alpha \), dictating how quickly the rolling motion accelerates.
Using the moment of inertia for a hollow cylinder, \( I = MR^2 \), and the no-slip condition that relates linear and angular accelerations \( a = R\alpha \), we can align the rotational equation, \( fR = MR^2\frac{a}{R} \). These dynamics give us insight into how the friction alone suffices to provide the required rotational movement, without additional forces.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. It plays a role similar to that of mass in linear motion.
For the thin-walled hollow cylinder like our lawn roller, the moment of inertia is calculated as \( I = MR^2 \). This value tells us how the mass is distributed further from the axis, making it more resistant to acceleration changes.
In solving our problem, the moment of inertia was crucial in determining how the torque produced by friction translates into angular acceleration. The equation \( I = MR^2 \) allows us to calculate the necessary rotational dynamics that couple with linear motion to keep the roller moving smoothly under the influence of the applied force and friction.