Question

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.

Short answer

Answer: The total energy of the rolling cylinder is 3375 J.

Step-by-step solution

Calculate the translational kinetic energy

The translational kinetic energy of the cylinder can be calculated using the formula: $$KE_{translational} = \frac{1}{2}mv^2$$ where m is the mass of the cylinder (5.00 kg) and v is the velocity of its center of mass (30.0 m/s). Plugging in the given values, $$KE_{translational} = \frac{1}{2}(5.00\,\text{kg})(30.0\,\text{m/s})^2$$ $$KE_{translational} = 2250\,\text{J}$$

Calculate the moment of inertia

Before we can calculate the rotational kinetic energy of the cylinder, we need to determine its moment of inertia, I. For a uniform solid cylinder, the moment of inertia around an axis passing through its center of mass and parallel to the axis of the cylinder is given by $$I = \frac{1}{2}mR^2$$ where m is the mass and R is the radius of the cylinder. However, we are not given the radius of the cylinder. But we do not need it since we are only calculating energy and the energy of the rolling cylinder only depends on its mass and velocity.

Calculate the rotational kinetic energy

To calculate the rotational kinetic energy of the cylinder, we use the formula: $$KE_{rotational} = \frac{1}{2}I\omega^2$$ The angular velocity ω of the cylinder can be related to its linear velocity v by $$v = R\omega$$ Rearranging and squaring gives $$\omega^2 = \frac{v^2}{R^2}$$ Substituting this in the KE_rotational formula, $$KE_{rotational} = \frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{v^2}{R^2}\right)$$ Simplifying, we obtain $$KE_{rotational} = \frac{1}{4}mv^2$$ Substituting the given values, $$KE_{rotational} = \frac{1}{4}(5.00\,\text{kg})(30.0\,\text{m/s})^2$$ $$KE_{rotational} = 1125\,\text{J}$$

Calculate the total kinetic energy

We can now calculate the total kinetic energy of the cylinder by summing the translational and rotational kinetic energies: $$KE_{total} = KE_{translational} + KE_{rotational}$$ $$KE_{total} = 2250\,\text{J} + 1125\,\text{J}$$ $$KE_{total} = 3375\,\text{J}$$ So the total energy of the rolling cylinder is 3375 J.

Key concepts

Translational Kinetic Energy

When we talk about translational kinetic energy, we're referring to the energy that an object has due to its motion through space. More specifically, it's the energy associated with the linear motion of an object's center of mass. For a rolling cylinder, its translational kinetic energy is crucial because it reflects how fast the cylinder is moving along a surface.

In our example with the solid cylinder, its translational kinetic energy is calculated using the fundamental formula:
\[KE_{translational} = \frac{1}{2}mv^2\]
Where:

• \(m\) is the mass of the cylinder
• \(v\) is the velocity of the center of mass

This part of kinetic energy plainly accounts for the object's bulk motion and not the rotation or any spinning movement.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation around a fixed axis. In the case of a rolling cylinder, this form of energy accounts for how fast the cylinder is spinning. It's an essential aspect of the total kinetic energy for objects that aren't just moving in a straight line but are also rotating, like wheels or planetary bodies.

To understand this concept, we use the formula:
\[KE_{rotational} = \frac{1}{2}I\omega^2\]
Here:

• \(I\) stands for the moment of inertia, which is different for various shapes and reflects how mass is distributed relative to the axis of rotation
• \(\omega\) is the angular velocity of the object

The cylinder's rotational kinetic energy, when combined with its translational kinetic energy, gives us a complete picture of its total energy while it's rolling.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to change in its rotation. It's akin to mass in the context of linear motion. Each object will have a unique moment of inertia depending on its mass distribution. For a rotating object, the further the mass is from the axis of rotation, the greater the moment of inertia will be.

For uniform solid cylinders, we apply the specific moment of inertia formula:
\[I = \frac{1}{2}mR^2\]
As noted in the cylinder energy problem, the moment of inertia depends on the mass (\(m\)) of the cylinder and the radius (\(R\)) from its axis of rotation. We use the moment of inertia to calculate the rotational kinetic energy, but when we don't have the radius, we can still glean essential information about the object's energy without it, based on its mass and velocity alone.

Uniform Solid Cylinder

A uniform solid cylinder is a three-dimensional object with two identical circular bases connected by a curved surface, where the mass is evenly distributed throughout its volume. For physics problems, we assume a uniform solid cylinder, like the one in our exercise, has a constant density, making calculations like kinetic energy and moment of inertia more straightforward.

These cylinders are prevalent in physics problems because they allow us to apply straightforward formulas for motion and energy, making them ideal for teaching these concepts. The uniformity means that for basic calculations, we can avoid complex integrals and focus on the essential aspects of translational and rotational motion.