Question

To illustrate the result (10.18), that the total KE of a body is just the rotational KE relative to any point that is instantaneously at rest, do the following: Write down the KE of a uniform wheel (mass M) rolling with speed \(v\) along a flat road, as the sum of the energies of the CM motion and the rotation about the CM. Now write it as the energy of the rotation about the instantaneous point of contact with the road and show that you get the same answer. (The energy of rotation is \(\frac{1}{2} I \omega^{2} .\) The moment of inertia of a uniform wheel about its center is \(I=\frac{1}{2} M R^{2} .\) That about a point on the rim is \(I^{\prime}=\frac{3}{2} M R^{2}\).)

Short answer

The total kinetic energy calculated from both perspectives is \( \frac{3}{4} Mv^2 \).

Step-by-step solution

Understanding the Kinetic Energy Components

The kinetic energy (KE) of a rolling wheel can be divided into translational KE due to the center of mass (CM) moving, and rotational KE due to the wheel spinning around its CM. The total KE is given by:\[KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2.\]Here, \(I = \frac{1}{2} MR^2\) is the rotational inertia and \(\omega\) is the angular velocity, \(\omega = \frac{v}{R}\) because rolling without slipping means speed \( v = \omega R \).

Calculating the CM Motion Energy

For the translational part of the kinetic energy due to CM motion:\[KE_{trans} = \frac{1}{2} Mv^2.\]No changes are needed, as it purely depends on the mass and linear speed of the wheel.

Calculating the Rotation about CM

The rotational kinetic energy about the CM is:\[KE_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{2} MR^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{4} Mv^2.\]This considers the wheel spinning around its center.

Total Kinetic Energy from CM Perspective

The total kinetic energy from the center of mass perspective is the sum of the translational and rotational components calculated:\[KE_{total} = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2.\]

Rotational Energy about Point of Contact

When considering the rotation about the instantaneous point of contact with the ground, we use the new rotational inertia about the point on the rim:\[I' = \frac{3}{2} MR^2.\]The rotational kinetic energy is:\[KE_{rot.contact} = \frac{1}{2} I' \omega^2 = \frac{1}{2} \left( \frac{3}{2} MR^2 \right) \left( \frac{v}{R} \right)^2 = \frac{3}{4} Mv^2.\]

Comparing Both Approaches

In both approaches, whether calculating from the CM motion and its rotation:\[KE_{total, CM} = \frac{3}{4} Mv^2,\]or from the energy of rotation about the point of contact:\[KE_{rot.contact} = \frac{3}{4} Mv^2,\]we find the same total kinetic energy. This illustrates that rolling and rotational kinetic energy computations are consistent irrespective of the reference point used.

Key concepts

Rotational Inertia

Rotational inertia, also known as the moment of inertia, is a measure of how much resistance an object shows to changes in its rotational motion. It depends on the object's mass distribution relative to the axis around which it is rotating.

For a wheel, the moment of inertia about its center of mass (CM) is denoted as \(I\) and is calculated using the formula \(I = \frac{1}{2} MR^2\). This formula tells us that half of the product of its mass \(M\) and the square of its radius \(R\) determines how it will resist angular acceleration.

When considering a different axis, like the point on the rim of the wheel, we use \(I'\), calculated as \(I' = \frac{3}{2} MR^2\). This increase happens because the mass is now distributed further from the new axis of rotation.

Understanding rotational inertia is key to predicting how objects will behave when forces are applied that might cause them to spin.

Center of Mass

The center of mass (CM) is a point in an object where its mass can be thought of as being concentrated. In the physics of rolling motion, this concept is crucial because it helps us break down the energy into translational and rotational components.

For a uniform wheel, the center of mass is right at its geometric center. When the wheel rolls on a surface without slipping, its CM moves along a straight path, giving rise to translational kinetic energy. This motion can be represented by the formula \(KE_{trans} = \frac{1}{2} Mv^2\), where \(v\) is the linear speed.

Simultaneously, the wheel rotates about its CM, contributing rotational kinetic energy, expressed as \(\frac{1}{4} Mv^2\) for its self-spin.

So, any change in the speed of the wheel will affect both the motion of its CM and its rotational energy.

Rolling Motion

Rolling motion combines both translational and rotational components. A wheel rolling on a surface without slipping is a perfect real-world example of this kind of motion. It's important to understand that rolling motion is a harmonious combination of moving forward while spinning around its CM.

In this scenario, the speed of the CM \(v\) relates directly to the wheel's angular speed \(\omega\) through \(v = \omega R\). This relationship shows how linear and rotational motions are interconnected.

• Translational kinetic energy accounts for the wheel's movement along the road.
• Rotational kinetic energy accounts for the wheel turning around its center of mass.

Interestingly, from a rotational perspective about the point of contact with the ground, the rotational inertia is higher \(I' = \frac{3}{2} MR^2\), but the total kinetic energy calculates to the same value \(\frac{3}{4} Mv^2\) as when analyzed from the CM.Understanding rolling motion is essential because it shows us how energy conservation and efficiency play out in everyday scenarios, such as wheels of vehicles moving.