Question

A uniform solid sphere of radius \(R,\) 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.

Short answer

Answer: The fraction of the sphere's total kinetic energy that is attributable to rotation is 1/3 or approximately 33.33%.

Step-by-step solution

Calculate the translational kinetic energy of the sphere

To find the translational kinetic energy (TKE) of the sphere, we will use the formula: TKE = (1/2)mv^2 where m is the mass and v is the linear velocity of the sphere.

Calculate the rotational kinetic energy of the sphere

To find the rotational kinetic energy (RKE) of the sphere, we will use the formula: RKE = (1/2)Iω^2 where I is the moment of inertia (I= (2/5)MR^2) and ω is the angular velocity of the sphere. Since the sphere is rolling without slipping, we can relate the linear velocity and the angular velocity using the equation: v = ωR

Substitute v = ωR in the rotational kinetic energy formula

RKE = (1/2)Iω^2 Now substitute v = ωR: RKE = (1/2)((2/5)MR^2)(v/R)^2

Simplify the rotational kinetic energy formula

Simplify the equation above: RKE = (1/2)((2/5)MR^2)(v^2/R^2) RKE = (1/5)Mv^2

Calculate the total kinetic energy of the sphere

The total kinetic energy (TKE_total) is the sum of the translational and rotational kinetic energies: TKE_total = TKE + RKE TKE_total = (1/2)Mv^2 + (1/5)Mv^2

Calculate the fraction of rotational kinetic energy

Finally, we will calculate the fraction of the sphere's total kinetic energy that is due to rotation. To do this, we will divide the rotational kinetic energy by the total kinetic energy: Fraction = RKE / TKE_total Fraction = ((1/5)Mv^2) / ((1/2)Mv^2 + (1/5)Mv^2) We can simplify this expression by canceling the Mv^2 terms: Fraction = (1/5) / (1/2 + 1/5)

Simplify the fraction

To simplify the fraction, we need to find a common denominator: Fraction = (1/5) / (2/5 + 1/5) Fraction = (1/5) / (3/5) Now, divide the numerators and denominators: Fraction = (1) / (3) There you have it! The fraction of the sphere's total kinetic energy that is attributable to rotation is 1/3 or approximately 33.33%.

Key concepts

Translational Kinetic Energy

Imagine pushing a toy car across a floor. The energy it carries as it moves, due solely to its motion, is what physicists call translational kinetic energy (TKE). It's the energy associated with the object's center of mass moving from one location to another.

For a solid sphere, or any object, TKE is calculated using the formula:
\[\begin{equation}TKE = \frac{1}{2}mv^2\end{equation}\]where m is the mass of the sphere, and v its velocity. This part of the kinetic energy is straightforward—just consider how fast the center of mass of the sphere is moving, without worrying about any spinning or rotating it might be doing.

In our example of a rolling sphere, the translational kinetic energy is the energy carrying the sphere straight down the path. It makes up a part of the sphere's total kinetic energy and can be understood as the energy you would calculate if the sphere slid across the surface without rolling.

Rotational Kinetic Energy

Now let's think about a spinning top. The energy that top needs to keep spinning on the spot is its rotational kinetic energy (RKE). Unlike TKE, RKE depends on the object's rotation around its center of mass.

For objects rotating around an axis, we calculate RKE using the formula:
\[\begin{equation}RKE = \frac{1}{2}I\omega^2\end{equation}\]with I being the moment of inertia (a measure of how mass is distributed in relation to the axis of rotation) and \omega being the angular velocity or how fast the object is spinning.

In our exercise, a sphere's rolling action is a combination of moving forward (translational movement) and spinning (rotational movement). Due to this dual movement, the sphere has both TKE and RKE. Importantly, because the sphere rolls without slipping, there's a direct relationship between its linear velocity and its rotational velocity.

Translation to Rotation: When an object rolls without slipping, its linear velocity v is equal to the product of its angular velocity \omega and its radius R:
\[\begin{equation}v = \omega R\end{equation}\]This relationship ensures that every point on the rolling sphere moves with the same speed as the sphere's center of mass in a translational sense.

Moment of Inertia

The concept of moment of inertia (I) can be a bit trickier to visualize. Think of it as the rotational analog to mass in linear motion. Just as mass influences how much an object resists changes in its linear velocity, moment of inertia determines how much it resists changes in its rotational velocity.

In our rolling sphere problem, the moment of inertia for a uniform solid sphere is given by:
\[\begin{equation}I = \frac{2}{5} M R^{2}\end{equation}\]Here, M represents the mass of the sphere, and R the radius. This formula tells us how the mass of the sphere is distributed in relation to its center.

Why is moment of inertia important in understanding the kinetic energy of rolling objects? Because it directly influences the RKE of the object. A higher moment of inertia means that the object will have more RKE for the same angular velocity. This is because a larger moment of inertia means the mass is spread out further from the axis of rotation, requiring more energy to spin. For our sphere, since the mass is spread uniformly, the factor of \frac{2}{5} is used in calculating its moment of inertia and thus its rotational kinetic energy.

Connecting Inertia and Energy: For our sphere, the RKE is five times less than the TKE due to the specific distribution of mass expressed by its moment of inertia. Knowing this allows us to understand and calculate the proportion of energy invested in the sphere's rotation compared to its straightforward, translational motion.