Question

A cylinder with mass \(M\) and radius \(R\) is rolling without slipping through a distance \(s\) along an inclined plane that makes an angle \(\theta\) with respect to the horizontal. Calculate the work done by (a) gravity, (b) the normal force, and (c) the frictional force.

Short answer

Question: Calculate the work done by gravity, normal force, and frictional force on a cylinder rolling without slipping down an inclined plane. Answer: The work done by gravity is \(Mgs\sin{\theta}\), the work done by the normal force is \(0\), and the work done by the frictional force is \(\frac{2}{3}Mgs\sin{\theta}\).

Step-by-step solution

Analyze forces acting on the cylinder

We are dealing with three forces acting on the cylinder: gravity (\(F_g\)), normal force (\(F_n\)), and frictional force (\(F_f\)). Notice that the force of gravity acts vertically downward, the normal force acts perpendicular to the inclined plane, and the frictional force acts parallel to the inclined plane opposing the motion.

Resolve gravitational force into components

Since the gravitational force acts vertically, it's necessary to resolve it into two components: one acting parallel to the inclined plane (\(F_{g\parallel}\)) and one acting perpendicular to the inclined plane (\(F_{g\perp}\)). The parallel component is given by \(F_{g\parallel} = Mg\sin{\theta}\), and the perpendicular component is given by \(F_{g\perp} = Mg\cos{\theta}\).

Calculating the normal force

Since the cylinder does not penetrate or leave the inclined plane, we can say that there is no vertical acceleration; thus the force acting on the cylinder perpendicular to the inclined plane should cancel out. This means that the normal force acting on the cylinder is equal in magnitude to the perpendicular component of gravity, i.e., \(F_n = F_{g\perp} = Mg\cos{\theta}\).

Calculating the frictional force

Since the cylinder is rolling without slipping, it is undergoing pure rolling motion, which means that the frictional force is static in nature. The frictional force is equal to the moment of inertia multiplied by the angular acceleration, divided by the radius (\(F_f = \frac{I \alpha}{R}\)). With a little more investigation, we can find that \(F_f = \frac{2}{3}Mg\sin{\theta}\) for a solid cylinder.

Calculating the work done by each force

(a) To calculate the work done by gravity, we need to consider only the component that is acting parallel to the inclined plane. The work done by gravity is given by \(W_g = F_{g\parallel} \cdot s = (Mg\sin{\theta}) \cdot s = Mgs\sin{\theta}\). (b) The work done by the normal force is zero, as the normal force is always perpendicular to the displacement. Since the angle between the normal force and the displacement is 90 degrees, the work done is \(W_n = F_n \cdot s \cdot \cos{90^\circ} = 0\). (c) The work done by the frictional force is given by \(W_f = F_f \cdot s = (\frac{2}{3}Mg\sin{\theta}) \cdot s = \frac{2}{3}Mgs\sin{\theta}\). So, the work done by gravity is \(Mgs\sin{\theta}\), the work done by the normal force is \(0\), and the work done by the frictional force is \(\frac{2}{3}Mgs\sin{\theta}\).

Key concepts

Rolling Motion

Rolling motion, also known as pure rolling, occurs when an object like a cylinder or a sphere moves in such a way that every point on its edge makes contact with the surface, turns once, and then lifts off again. During rolling motion, the object does not slip, meaning the velocity at the point of contact is momentarily zero relative to the surface upon which it rolls. Specifically for a cylinder, the condition of rolling without slipping implies a direct relationship between the linear velocity of the cylinder's center of mass and its angular velocity.

This can be mathematically expressed as: \( v = \omega R \), where \( v \) is the linear velocity, \( \omega \) is the angular velocity, and \( R \) is the radius of the cylinder. Due to this condition, the frictional force encountered by the rolling object is static friction, which ensures that the point of contact is at rest with respect to the surface. Consequently, the static friction has no 'slip work' associated with it and hence does not dissipate energy as it would if the object was sliding. The calculation in the solution reflects this: work done by friction during rolling without slipping is not zero due to the rotation of the cylinder.

Inclined Plane

An inclined plane, a flat surface tilted at an angle, significantly influences the forces acting on an object placed on it. When dealing with problems such as a cylinder rolling down an inclined plane, it's essential to understand how forces change with respect to the incline. On a flat surface, the force due to gravity acts straight down on the object, and no motion occurs as there is no component of this force causing horizontal displacement. However, on an inclined plane, gravity can be resolved into two vector components: one perpendicular to the slope and one parallel to it. The normal force, exerted by the surface on the object, counteracts only the perpendicular component of gravity, leaving the parallel component unbalanced, causing the object to accelerate down the slope.

The angle of the incline (\( \theta \)) dictates the magnitude of these components, with steeper angles leading to greater parallel forces, hence faster acceleration. This has direct implications on the work calculations. Since only the parallel component of gravity (\( F_{g\rVert} \)) does work on the cylinder as it rolls down the inclined plane, the inclination angle is crucial to determining the total work done by gravity.

Gravitational Force Components

The force of gravity is a key player in physics problems involving motion on Earth. It's a conservative force, meaning the work done by gravity on an object moving between two points does not depend on the path taken. However, when working with inclined planes, it’s essential to decompose the gravitational force into components that align with the motion's geometry.

This decomposition results in a parallel component (\( F_{g\rallel} \)) causing the object to move down the slope and a perpendicular component (\( F_{g\rapidargout} \)) which is counterbalanced by the normal force. The parallel component can be calculated using the sine function (\( F_{g\rallel} = Mg\sin{\theta} \)), where \( M \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the inclined plane.

Understanding these components is crucial for determining the work done by gravity, as only the parallel component contributes to work. Work by gravity, as given in the solution, is the product of this parallel component and the displacement along the inclined plane (\( W_g = Mgs\sin{\theta} \)). It is important to note that the perpendicular component of gravity does no work since it does not cause displacement.