When a sphere of moment of inertia I rolls down on an inclined plane the percentage of total energy which is rotational, is approximately

When a sphere rolls down an inclined plane, it possesses both translational and rotational kinetic energy. To determine the percentage of the total energy that is rotational, we need to analyze the energy distribution between these two forms of motion. Let's break this down step by step.

Understanding the Energy Components

As the sphere rolls down the incline, it converts gravitational potential energy into kinetic energy. This kinetic energy can be divided into two parts:

• Translational Kinetic Energy (TKE): This is the energy due to the motion of the center of mass of the sphere, given by the formula:

TKE = (1/2)mv²

• Rotational Kinetic Energy (RKE): This is the energy due to the rotation of the sphere around its center of mass, expressed as:

RKE = (1/2)Iω²

Relating Linear and Angular Quantities

For a sphere rolling without slipping, there is a relationship between the linear velocity (v) of the center of mass and the angular velocity (ω) of the sphere:

v = rω

Here, r is the radius of the sphere. The moment of inertia (I) for a solid sphere is given by:

I = (2/5)mr²

Calculating Total Energy

When the sphere rolls down the incline, the total mechanical energy (E_total) at any point can be expressed as:

E_total = TKE + RKE

Substituting the expressions for TKE and RKE, we have:

E_total = (1/2)mv² + (1/2)Iω²

Now, substituting I and using the relationship between v and ω:

E_total = (1/2)mv² + (1/2)((2/5)mr²)(v/r)²

E_total = (1/2)mv² + (1/5)mv²

E_total = (1/2 + 1/5)mv² = (5/10 + 2/10)mv² = (7/10)mv²

Finding the Percentage of Rotational Energy

Now, we can find the percentage of the total energy that is rotational:

RKE = (1/5)mv²

Percentage of rotational energy = (RKE / E_total) × 100

Percentage of rotational energy = [(1/5)mv² / (7/10)mv²] × 100

Notice that the mass (m) and velocity squared (v²) cancel out:

Percentage of rotational energy = [(1/5) / (7/10)] × 100 = (2/7) × 100 ≈ 28.57%

Final Thoughts

Thus, when a solid sphere rolls down an inclined plane, approximately 28.57% of the total energy is rotational. This analysis highlights the interplay between translational and rotational motion, showcasing how energy is conserved and transformed in a rolling object.