Question

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

Short answer

The short answer is: The angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\).

Step-by-step solution

Find the total potential energy at the incline's top.

At the top of the inclined plane, the cylinder has a potential energy given by: \(PE = mgh\) Where: - \(PE\) = potential energy of the cylinder - \(m\) = mass of the cylinder (\(M\)) - \(g\) = acceleration due to gravity - \(h\) = height of the inclined plane

Find the rotational and translational kinetic energies at the bottom of the incline.

At the bottom of the inclined plane, the cylinder has both translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation). These can be given as follows: \(KE_t = \frac{1}{2} mv^2\) \(KE_r = \frac{1}{2} I \omega^2\) Where: - \(KE_t\) = translational kinetic energy - \(KE_r\) = rotational kinetic energy - \(v\) = linear velocity at the bottom of the incline - \(I\) = moment of inertia of the cylinder - \(\omega\) = angular velocity of the cylinder at the bottom of the incline The moment of inertia for a solid cylinder is \(I = \frac{1}{2} MR^2\).

Apply the conservation of energy.

By the conservation of energy, the potential energy at the top of the incline is equal to the sum of the translational and rotational kinetic energies at the bottom. Thus, \(mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\) Substituting the moment of inertia for the solid cylinder, we get: \(mgh = \frac{1}{2} mv^2 + \frac{1}{4} MR^2 \omega^2\)

Relate linear velocity and angular velocity.

For a rolling object without slipping, the linear velocity \(v\) and the angular velocity \(\omega\) are related as follows: \(v = \omega R\) We can substitute this into the conservation of energy equation: \(mgh = \frac{1}{2} m(\omega R)^2 + \frac{1}{4} MR^2 \omega^2\)

Solve for angular velocity.

Now we can solve for the angular velocity \(\omega\): \(\omega^2(R^2 + \frac{1}{2}R^2) = 2gh\) Solving for \(\omega\), we get: \(\omega = \sqrt{\frac{2gh}{\frac{3}{2}R^2}}\) \(\omega = \sqrt{\frac{4gh}{3R^2}}\) \(\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}\) Thus, the angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\), which corresponds to option C.

Key concepts

Angular Velocity

Angular Velocity is a measure of how quickly an object is rotating around its axis. It tells us the speed of rotation and the direction of the axis of rotation.
For the solid cylinder rolling down the inclined plane, its angular velocity changes as it descends. When it reaches the bottom, this angular velocity \(\omega\) can be calculated using energy conservation and kinematic relationships.

• The angular velocity of an object is key to understanding rotational motion.
• The formula for angular velocity at the bottom of the incline involves both the gravitational potential energy that converted into kinetic energy and how that energy is distributed between rotational and translational movement.
• Higher angular velocity means that the cylinder spins faster.

Linking this concept closely with the problem, the cylinder's angular velocity at the bottom is determined by the relationship \(\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}\). This result comes from balancing the initial potential energy with the energy forms at the bottom of the incline.

Conservation of Energy

The Conservation of Energy principle states that energy in a closed system is constant, meaning it can't be created or destroyed but can change forms. This principle is crucial in understanding rolling motion down an inclined plane.
In the given exercise, when the cylinder is at the top of the incline, it has maximum gravitational potential energy given by \(PE = mgh\). As it rolls down, this energy transforms into kinetic energy, split between translational kinetic energy and rotational kinetic energy.

• The total energy at the top is equal to the total energy at the bottom.
• Energy conservation allows us to derive the missing quantities using known variables.

By setting the initial potential energy equal to the sum of translational and rotational kinetic energies at the bottom, we can find the angular velocity. We demonstrated this by deriving the equation \(mgh = \frac{1}{2} mv^2 + \frac{1}{4} MR^2 \omega^2\).

Moment of Inertia

Moment of Inertia is a measure of an object's resistance to changes in its rotation. It plays a similar role in angular motion as mass does in linear motion.
In this exercise, the moment of inertia for a solid cylinder is expressed as \(I = \frac{1}{2} MR^2\). This equation is crucial for calculating the rotational kinetic energy of the cylinder.

• It depends on both the mass and the radius of the cylinder.
• Higher moment of inertia implies more resistance to changes in rotational speed.
• It's important for understanding how energy distributes between translational and rotational motion.

The moment of inertia directly affects the calculation of rotational kinetic energy, which plays a part in determining the cylinder's angular velocity when reaching the bottom of the inclined plane.

Inclined Plane

An Inclined Plane is a flat surface tilted at an angle to the horizontal. This setup introduces gravitational potential energy and influences the motion of rolling objects.
In our exercise, the cylinder rolls down this plane, gaining speed due to gravity. The height \(h\) of the incline plays a critical role as it affects the potential energy that converts into kinetic energy.

• The steeper the inclined plane, the greater the acceleration of the object.
• The height of the plane directly affects the potential energy available for conversion into kinetic energies.
• The incline angle does not appear explicitly, but it impacts the gravitational force component acting along the plane which accelerates the cylinder.

In essence, the inclined plane enables us to analyze how gravitational potential energy transforms into motion-driven kinetic energies, both translational and rotational.

Rotational Kinetic Energy

Rotational Kinetic Energy is the energy due to an object's rotation and is calculated with \(KE_r = \frac{1}{2} I \omega^2\). It is a critical component when analyzing objects rolling down an inclined plane.
For the cylinder in this exercise, as it descends the slope, its rotational speed increases, which increases its rotational kinetic energy.

• It contributes to the total kinetic energy the cylinder has at the bottom of the inclined plane.
• Understanding rotational kinetic energy helps in connecting the moment of inertia and angular velocity with physical motion.
• It's a portion of the energy that initially started as gravitational potential energy at the top.

By recognizing how kinetic energy splits between translational and rotational forms, you can solve for the cylinder's angular velocity, a critical parameter in understanding its motion upon reaching the bottom.