To determine how far up the inclined plane the ring travels before rolling back down, we can apply the principles of energy conservation. The key here is to consider both the translational and rotational kinetic energy of the ring, as well as the potential energy it gains as it ascends the incline.

Understanding the Energy Components

The total mechanical energy of the ring at any point consists of its kinetic energy (both translational and rotational) and gravitational potential energy. The equations we will use are:

• Kinetic Energy (Translational): \( KE_{trans} = \frac{1}{2} mv^2 \)
• Kinetic Energy (Rotational): \( KE_{rot} = \frac{1}{2} I \omega^2 \)
• Potential Energy: \( PE = mgh \)

Where:

• m = mass of the ring (2.40 kg)
• v = speed of the ring (2.80 m/s)
• I = moment of inertia of the ring
• ω = angular velocity
• h = height gained on the incline

Calculating the Moment of Inertia

The moment of inertia \( I \) for a ring about its central axis is given by:

\( I = m \left( \frac{R_{outer}^2 + R_{inner}^2}{2} \right) \)

Substituting the values:

\( I = 2.40 \, \text{kg} \left( \frac{(0.08 \, \text{m})^2 + (0.06 \, \text{m})^2}{2} \right) = 2.40 \, \text{kg} \left( \frac{0.0064 + 0.0036}{2} \right) = 2.40 \, \text{kg} \left( \frac{0.0100}{2} \right) = 2.40 \, \text{kg} \times 0.005 = 0.012 \, \text{kg m}^2 \)

Finding Angular Velocity

Since the ring rolls without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:

\( v = R \omega \)

Where \( R \) is the average radius. For our ring, we can use the average of the inner and outer radius:

\( R = \frac{R_{outer} + R_{inner}}{2} = \frac{0.08 + 0.06}{2} = 0.07 \, \text{m} \)

Thus, \( \omega = \frac{v}{R} = \frac{2.80}{0.07} = 40 \, \text{rad/s} \)

Calculating Kinetic Energies

Now we can calculate the total kinetic energy at the initial position:

• Translational: \( KE_{trans} = \frac{1}{2} (2.40) (2.80)^2 = \frac{1}{2} (2.40) (7.84) = 9.408 \, \text{J} \)
• Rotational: \( KE_{rot} = \frac{1}{2} (0.012) (40)^2 = \frac{1}{2} (0.012) (1600) = 9.6 \, \text{J} \)

Total initial kinetic energy \( KE_{total} = KE_{trans} + KE_{rot} = 9.408 + 9.6 = 19.008 \, \text{J} \)

Potential Energy at Maximum Height

As the ring ascends the incline, it converts kinetic energy into potential energy. The height \( h \) gained can be related to the distance \( d \) traveled up the incline using the angle of the incline:

\( h = d \sin(\theta) \)

At the maximum height, all kinetic energy will be converted into potential energy:

\( KE_{total} = PE \)

\( 19.008 = mgh = (2.40)(9.81)(d \sin(36.9^\circ)) \)

Solving for Distance Up the Incline

First, calculate \( \sin(36.9^\circ) \approx 0.6 \):

Now substituting into the equation:

\( 19.008 = (2.40)(9.81)(d)(0.6) \)

\( 19.008 = 14.118(d) \)

\( d = \frac{19.008}{14.118} \approx 1.34 \, \text{m} \)

Thus, the ring travels approximately 1.34 meters up the incline before rolling back down.