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

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.

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

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 \(v\): \(v = \sqrt{\frac{4gh}{3}}\)

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: \(v = \sqrt{\frac{4gh}{3}}\) \(a =\frac{2}{3}g\sin\theta\)