Problem 17 - University Physics with Modern Physics

Question

A uniform solid sphere of radius \(R, \operatorname{mass} M,\) and moment of inertia \(I=\frac{2}{5} M R^{2}\) is rolling without slipping along a horizontal surface. Its total kinetic energy is the sum of the energies associated with translation of the center of mass and rotation about the center of mass. Find the fraction of the sphere's total kinetic energy that is attributable to rotation.

Step-by-Step Solution

Answer

The fraction of the sphere's total kinetic energy that is due to rotation is 2/7.

Step 1: Find the translational kinetic energy

To find the translational kinetic energy, we use the formula: \(T = \frac{1}{2}mv^2\) where \(m\) is the mass of the sphere, and \(v\) is its linear velocity.

Step 2: Find the rotational kinetic energy

To find the rotational kinetic energy, we use the formula: \(R = \frac{1}{2}I\omega^2\) where \(I\) is the moment of inertia of the sphere, and \(\omega\) is its angular velocity.

Step 3: Relate the linear and angular velocities

Since the sphere is rolling without slipping, we can relate the linear velocity \(v\) to its angular velocity \(\omega\): \(v = R\omega\)

Step 4: Express the rotational kinetic energy in terms of linear velocity

Substitute the relation between linear and angular velocities into the expression for rotational kinetic energy: \(R = \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{5}mv^2\)

Step 5: Find the total kinetic energy

The total kinetic energy is the sum of translational and rotational kinetic energies: \(K = T + R = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\)

Step 6: Calculate the fraction of rotational kinetic energy in the total kinetic energy

Finally, we find the fraction of the rotational kinetic energy in the total kinetic energy: \(\frac{R}{K} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5}\cdot\frac{10}{7} = \boxed{\frac{2}{7}}\) So, the fraction of the sphere's total kinetic energy that is attributable to rotation is \(\frac{2}{7}\).

Key Concepts

Translational Kinetic Energy

Translational kinetic energy refers to the energy associated with the motion of an object's center of mass as it travels through space. For a solid sphere rolling on a flat surface, this motion involves every point on the sphere moving forward in a straight line.
To calculate this energy, we use the formula:

• Translational kinetic energy, \(T = \frac{1}{2}mv^2\)

Here, \(m\) represents the mass of the sphere, and \(v\) is the linear velocity of its center of mass. Remember, this type of kinetic energy applies to any object moving in a straight line, not just spheres. In this scenario, half of the total kinetic energy is due to this straightforward motion of the object itself.

Rotational Kinetic Energy

Rotational kinetic energy is related to the rotation of an object around its center of mass. In a rolling sphere, this type of energy accounts for how the object spins as it moves.
We calculate rotational kinetic energy using the formula:

• Rotational kinetic energy, \(R = \frac{1}{2}I\omega^2\)

Where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. The moment of inertia reflects how the mass is distributed with regard to the axis of rotation, affecting how easily the object spins. For a uniform solid sphere, the moment of inertia is expressed as \(I = \frac{2}{5} MR^2\), which is essential for calculating the rotational kinetic energy. This aspect of kinetic energy highlights the object's rotational motion as it moves forward.

Moment of Inertia

The moment of inertia is a crucial concept in rotational dynamics. It acts as a rotational equivalent of mass in linear motion, indicating how hard it is to change an object's rotational state.
The moment of inertia \(I\) of a body changes based on its shape and the axis it's rotating around. For a sphere, it's given by:

• \(I = \frac{2}{5} MR^2\).

This formula shows that the moment of inertia depends on both the mass \(M\) and the square of the radius \(R\) of the sphere. As the sphere's mass or radius increases, its resistance to change in rotational speed also increases. This is why it's easier to spin a small marble than a bowling ball; the latter has a larger moment of inertia due to its mass and size.

Rolling Motion

Rolling motion is a combination of translational and rotational motion, crucial for understanding how objects like wheels and spheres move. When an object rolls "without slipping," it means each point on the object touching the surface is momentarily at rest relative to the surface.
In rolling motion:

• The linear velocity \(v\) and angular velocity \(\omega\) are linked by the relationship \(v = R \omega\), where \(R\) is the object's radius.

This connection is key when calculating the total kinetic energy of the sphere, comprising both its translational and rotational kinetic energies. By understanding how these energies combine, we see why the motion appears smooth and continuous, like a ball rolling down a hill or a car tire on a road.