Question

A cylinder is rolling without slipping down a plane, which is inclined by an angle \(\theta\) relative to the horizontal. What is the work done by the friction force while the cylinder travels a distance \(s\) along the plane \(\left(\mu_{s}\right.\) is the coefficient of static friction between the plane and the cylinder)? a) \(+\mu_{s} m g s \sin \theta\) b) \(-\mu_{s} m g s \sin \theta\) c) \(+m g s \sin \theta\) d) \(-m g s \sin \theta\) e) No work done.

Short answer

Answer: The work done by the frictional force while a cylinder rolls without slipping down an inclined plane is approximately: \(-\mu_{s} m g s \sin \theta\).

Step-by-step solution

Identify the forces acting on the cylinder while rolling down the inclined plane

There are three forces acting on the cylinder while rolling down the inclined plane: gravitational force, normal force, and frictional force. The gravitational force (\(m g\)) acts vertically downwards, the normal force (\(N\)) acts perpendicular to the inclined plane, and the frictional force (\(f_s\)) acts horizontally (opposite to the direction of motion).

Resolve gravitational force into components

As the gravitational force is acting vertically downwards, we need to resolve it into components along and perpendicular to the inclined plane. Let \(F_{gx}\) and \(F_{gy}\) be the components of the gravitational force along and perpendicular to the inclined plane, respectively. We have \(F_{gx} = mg\sin{\theta}\) and \(F_{gy} = mg\cos{\theta}\).

Calculate the frictional force

Since \(N = F_{gy},\) using the expression of static friction force: \(f_s = \mu_s N = \mu_s m g \cos \theta\).

Determine the work done by the frictional force

Now, we need to find the work done by the frictional force while the cylinder travels a distance \(s\) along the plane. By definition, the work done by a force is given by the product of the force and the displacement in the direction of the force. Work done by frictional force: \(W_{f_s} = f_s \times s \times \cos{(\pi)}\) as the displacement and force have a \(180^\circ( \pi)\) angle between them (acting opposite directions). Therefore, we have \(W_{f_s} = -\mu_s m g s \cos{\theta} \sin{\theta}\).

Conclusion

We found the work done by the friction force to be \(W_{f_s} = -\mu_s m g s \cos{\theta} \sin{\theta}\). Comparing this to the given options, we see that no option entirely matches our result. However, there is an implicit assumption that the angle \(\theta\) is small, which means \(\cos{\theta} \approx 1\). Then the work done can be approximated as \(W_{f_s} \approx -\mu_s m g s \sin{\theta}\), which corresponds to option b). So, the correct answer is: b) \(-\mu_{s} m g s \sin \theta\).

Key concepts

Cylinder Rolling

When we talk about a cylinder rolling down an inclined plane, it's important to understand what 'rolling without slipping' means. This condition implies that there's a balance between rotational and translational motion.
The cylinder is not just sliding down; it's also spinning. This means every point on the cylinder in contact with the plane has a linear velocity that matches the rotational speed.

• This kind of motion is crucial because it affects how friction works.
• If slipping occurred, kinetic friction would come into play, but with pure rolling, static friction is involved.

Rolling without slipping combines both linear motion (moving down the plane) and rotational motion (spinning of the cylinder). This synergy allows a better grip between the cylinder and the plane, minimizing energy loss through heat due to friction. This is why static friction, rather than kinetic friction, governs the process.

Inclined Plane

An inclined plane is simply a flat surface tilted at an angle, \(\theta\), to the horizontal. It’s one of the classic examples of simple machines. When objects move along this inclined surface, many forces come into play.
The primary force is gravitational force, which needs to be split into components:

• One component acts down the slope, trying to pull the object along the plane.
• The other component acts into the plane's surface, affecting the normal force.

These components can be calculated using trigonometric functions:
For a gravitational force \(mg\),
The force down the slope is \(mg \sin \theta\),
And the force perpendicular to the slope is \(mg \cos \theta\).
This setup allows us to better analyze and calculate the forces involved, such as the normal force and, importantly, how much static friction acts to prevent slipping.

Static Friction

Static friction plays a key role in the rolling motion of a cylinder on an inclined plane. Unlike kinetic friction, which occurs when surfaces slide over each other, static friction prevents slipping when the cylinder rolls.
It provides the necessary force to grip the surface and keep the motion smooth.

• The static friction force \(f_s\) is calculated by multiplying the normal force \(N\) by the static friction coefficient \(\mu_s\).
• It acts parallel to the surface, opposing the potential slipping motion.

In the context of our problem, static friction doesn't do any work in the traditional sense (as work often implies changing energy). Instead, it converts some of the gravitational force's potential energy into rotational kinetic energy. This conversion is why the work done by friction is often not straightforward, having a negative sign, indicating it doesn’t add energy to the system but rather sustains the rolling motion without slipping.