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 that makes an angle \(\theta\) with respect to the horizontal.

Short answer

The final speed of the cylindrical object is given by \(v_f = \sqrt{\frac{4g(s\sin{\theta})}{3}}\), and the acceleration of the cylinder along the inclined plane is given by \(a = \frac{2g\sin{\theta}}{3}\).

Step-by-step solution

Compute the initial and final mechanical energy of the system

Initially, the cylindrical object is at rest, so it only has gravitational potential energy (PE) due to its height. We can find the initial height of the cylinder with respect to the inclined plane using the equation \(h = s\sin{\theta}\). The initial potential energy is given by \(PE_i = Mgh\), where \(g\) is the acceleration due to gravity. After rolling down the inclined plane, the object has both kinetic energy due to its translational motion and kinetic energy due to its rotation. The final total kinetic energy is given by the sum of both translational and rotational kinetic energies: \(KE_f = \frac{1}{2}Mv_f^2 + \frac{1}{2}I\omega^2\), where \(v_f\) is the final linear speed of the cylinder, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. As the object rolls without slipping, we can relate the linear and angular velocities by the equation \(v_f = R\omega\).

Calculate the moment of inertia of the cylinder

For a solid cylinder rolling without slipping, the moment of inertia is given by the equation \(I = \frac{1}{2}MR^2\).

Apply the conservation of mechanical energy principle

Using the conservation of mechanical energy principle, we can set the initial potential energy equal to the final kinetic energy, which gives us the equation \(PE_i = KE_f\) or \(Mgh = \frac{1}{2}Mv_f^2 + \frac{1}{2}I\omega^2\).

Find the final velocity of the cylinder#EndTitle#By substituting \(I = \frac{1}{2}MR^2\) and \(v_f = R\omega\), we get the equation \(Mgh = \frac{1}{2}Mv_f^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v_f^2}{R^2})\). Solving for \(v_f\), we obtain \(v_f^2 = \frac{4gh}{3}\), so \(v_f = \sqrt{\frac{4gh}{3}}\). Thus, the final velocity is given by \(v_f = \sqrt{\frac{4g(s\sin{\theta})}{3}}\).

Step 5: Calculate the acceleration of the cylinder along the inclined plane#EndTitle#We will use the kinematic equation \(v_f^2 = v_i^2 + 2as\), where \(v_i\) is the initial velocity (which is 0 in this case), and \(a\) is the acceleration we want to find. Plugging in the values, we get \(\frac{4gh}{3} = 2as \Rightarrow a = \frac{2gh}{3s}\). Since \(h = s\sin{\theta}\), we can rewrite this as \(a = \frac{2g\sin{\theta}}{3}\). Therefore, the acceleration of the cylinder along the inclined plane is given by \(a = \frac{2g\sin{\theta}}{3}\).

Key concepts

Moment of Inertia

Understanding the moment of inertia is crucial when analyzing the motion of rotating bodies. Essentially, it's the rotational equivalent of mass for linear motion. It tells us how difficult it is to change the rotational speed of an object. For a solid cylinder, as in our example, the moment of inertia is given by the formula \( I = \frac{1}{2}MR^2 \), where \( M \) is the mass and \( R \) is the radius of the cylinder.

Different shapes have different moments of inertia because their mass is distributed differently relative to the axis of rotation. This concept is paramount when assessing the cylinder's energy on the inclined plane since it impacts the rotational kinetic energy, which is part of the total mechanical energy conserved during its descent.

When an object rolls without slipping, like the cylinder down the inclined plane, a direct relationship exists between its linear and angular velocities. This coupling is essential for the cylinder's energy calculation, as any change in linear motion will directly affect the rotational movement and vice versa.

Kinematic Equations

Kinematic equations are the tools that allow us to describe the motion of objects with respect to time. They are particularly helpful when analyzing straight-line motion with uniform acceleration, like the cylinder rolling down the inclined plane.

One of the key equations used in our problem is \( v_f^2 = v_i^2 + 2as \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance covered. Since the cylinder starts from rest, \( v_i = 0 \), which simplifies our equation considerably. By using this kinematic equation, we can solve for the acceleration of the cylinder as it rolls down the plane.

This equation is crucial for connecting the distance traveled by the cylinder and its acceleration to find its velocity. Using this information, we were able to calculate the final velocity and, consequently, the kinetic energy part of the conservation of mechanical energy equation.

Inclined Plane Physics

An inclined plane is a flat surface tilted at an angle, like a ramp, which makes analysis of motion more complex than movement along a horizontal surface. The force of gravity on the object is divided into two components: one perpendicular to the plane and one parallel to it. It's this parallel component that accelerates the object down the slope.

The steeper the plane (larger the angle \( \theta \)), the greater the acceleration of an object sliding or rolling down it. Our problem deals with a cylindrical object on an inclined plane, where gravity induces both translation (down the slope) and rotation (around the cylinder's axis).

The physics of the inclined plane also involves the no-slip condition, indicating that the object rolls purely without sliding. This results in a constant ratio between the object's angular speed and its linear velocity, implying a direct relationship needed to analyze the motion comprehensively and apply the law of conservation of mechanical energy effectively.