Question

Using the conservation of mechanical energy, calculate the final speed and the acceleration of a cylindrical object of mass \(M\) and radius \(R\) after it rolls a distance \(s\) without slipping along an inclined plane of angle \(\theta\) with respect to the horizontal

Short answer

Answer: The final speed and acceleration of the cylindrical object after rolling down an inclined plane without slipping are: \(v = \sqrt{\frac{4gh}{3}}\) \(a =\frac{2}{3}g\sin\theta\)

Step-by-step solution

Determine the initial and final mechanical energies of the system

Initially, the cylindrical object is at rest at the top of the inclined plane, and all its mechanical energy is in the form of potential energy. At the bottom, its potential energy is converted into rotational and translational kinetic energies. Write the initial and final mechanical energies using the following equations (consider the potential energy reference at the bottom of the inclined plane): \(E_{initial} = Mgh\) \(E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\) Where \(h = s\sin\theta\) is the height the object rolls down, \(v\) is the linear speed at the bottom of the inclined plane, and \(I\) and \(\omega\) are the moment of inertia and angular velocity at the bottom, respectively.

Determine the relation between linear and angular velocities

Since the object rolls without slipping, we can relate the linear and angular velocities through the radius of the cylinder: \(v = R\omega\)

Solve for the final linear speed

Apply the conservation of mechanical energy: \(Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\) The moment of inertia of a cylinder of mass \(M\) and radius \(R\) is given by: \(I = \frac{1}{2}MR^2\) Replace \(\omega\) using the relation between linear and angular velocities and substitute the moment of inertia: \(Mgh = \frac{1}{2}Mv^2 + \frac{1}{2} \left(\frac{1}{2}MR^2\right) \left(\frac{v}{R}\right)^2\) Simplify the equation and solve for \(v\): \(v = \sqrt{\frac{4gh}{3}}\)

Calculate the acceleration down the inclined plane

From the conservation of energy, we can find the acceleration by considering the work-energy theorem. The work done on the object by gravitational force is: \(W = Mgh\) The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy: \(W = \Delta K = K_{final} - K_{initial}\) We know that the initial kinetic energy is 0, and the final kinetic energy is \(\frac{1}{2}Mv^2\). Thus, we have: \(Mgh = \frac{1}{2}Mv^2\) Cancel out the mass \(M\), and use the final speed we calculated in step 3: \(g\sin\theta = \frac{1}{2}\left(\frac{4gh}{3}\right)\) Solve for \(a\) (the acceleration down the inclined plane): \(a =\frac{2}{3}g\sin\theta\) The final speed and acceleration of the cylindrical object after rolling a distance \(s\) down the inclined plane are: \(v = \sqrt{\frac{4gh}{3}}\) \(a =\frac{2}{3}g\sin\theta\)

Key concepts

Mechanical Energy

Mechanical energy is a fundamental concept in physics, describing the energy due to the position and motion of an object. In this exercise, the principle of the conservation of mechanical energy is applied. This principle states that in a closed system where only conservative forces, like gravity, are acting, the total mechanical energy remains constant. It consists of two main forms:

• Potential Energy (PE), denoted by \( Mgh \), where \( M \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height.
• Kinetic Energy (KE), which can be further divided into translational and rotational kinetic energies. Translational kinetic energy is given by \( \frac{1}{2}Mv^2 \), and rotational kinetic energy can be written as \( \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

The exercise demonstrates how potential energy, at the top of the incline, is transformed into kinetic energy as the object rolls down, factoring in both translational and rotational components. This allows us to calculate the speed at the bottom using conservation equations.

Rotational Kinematics

Rotational kinematics deals with the motion of objects that rotate. For this problem, a critical connection is made between the linear velocity \( v \) and the angular velocity \( \omega \). For objects rolling without slipping, such as the cylinder in this exercise, this relationship is expressed as:

• \( v = R\omega \)

where \( R \) is the radius of the cylinder. This equation ensures that every point on the perimeter of the rolling object travels the same linear distance as the object itself, allowing us to calculate one variable when the other is known.
Understanding rotational kinematics is crucial to calculate the final speed of the cylinder using conservation of energy. The equation for the moment of inertia of a cylinder, \( I = \frac{1}{2}MR^2 \), plays a key role when substituted into the energy equation. This substitution, along with the relationship \( v = R\omega \), allows us to resolve the energies involved precisely.

Inclined Plane Dynamics

Inclined plane dynamics is an essential aspect when analyzing motions involving a slope. The physics of an inclined plane primarily involves decomposing forces and the resulting accelerations. In this example, the inclined plane is at an angle \( \theta \), which influences both the object's acceleration and its speed at the bottom.
As the cylinder rolls down, gravity does work on it, transferring potential energy into kinetic energy. The effective component of gravitational force acting along the inclined plane is \( Mg\sin\theta \), driving the object to accelerate as it rolls down. This component simplifies calculations, as it directly relates to the object's acceleration:

• \( a = \frac{2}{3}g\sin\theta \)

This reflects how the angle of inclination affects the cylinder's acceleration. The smaller the angle, the less acceleration; the steeper the angle, the more the acceleration. Understanding these dynamics is vital for solving problems involving any object moving along an incline.