A solid disk and a solid sphere of the same radius but possibly different masses roll without slipping down identical inclined planes of the same height and length. Which object will reach the bottom of the plane first?

Introduction / Context:
This question tests understanding of rotational motion and energy distribution in rolling bodies. When different objects roll down an inclined plane without slipping, their acceleration depends not only on gravity and the angle of the plane but also on how their mass is distributed, which is captured by the moment of inertia. Comparing a solid disk and a solid sphere of the same radius is a classic conceptual problem in mechanics that illustrates how rotational inertia affects translational motion.

Given Data / Assumptions:

• There are two rigid bodies: a solid disk and a solid sphere.
• Both have the same radius but may have different masses.
• They roll without slipping down inclined planes of the same height and length.
• Friction is sufficient only to prevent slipping; it does no net work.
• We neglect air resistance and assume uniform gravitational field.

Concept / Approach:
When a body rolls without slipping, gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy. The total acceleration of the body down the plane depends on its moment of inertia about the centre. For a solid disk of mass m and radius R, the moment of inertia is I = (1 / 2) * m * R^2. For a solid sphere, the moment of inertia is I = (2 / 5) * m * R^2. The expression for linear acceleration a of a rolling object on an incline is a = g * sin(theta) / (1 + I / (m * R^2)). A smaller ratio I / (m * R^2) gives a larger acceleration. Because the solid sphere has a smaller I / (m * R^2) than the solid disk, it experiences greater acceleration and reaches the bottom first. The result is independent of mass.

Step-by-Step Solution:
Step 1: Write the expression for the moment of inertia of a solid disk: I_disk = (1 / 2) * m * R^2. Step 2: Write the expression for the moment of inertia of a solid sphere: I_sphere = (2 / 5) * m * R^2. Step 3: Compute the ratio I / (m * R^2) for each shape. For the disk, I_disk / (m * R^2) = 1 / 2. For the sphere, I_sphere / (m * R^2) = 2 / 5. Step 4: Compare these ratios: 1 / 2 is 0.5 and 2 / 5 is 0.4, so the solid sphere has a smaller I / (m * R^2). Step 5: Use the formula for linear acceleration of a rolling object on an incline: a = g * sin(theta) / (1 + I / (m * R^2)). A smaller value of I / (m * R^2) gives a larger acceleration. Step 6: Therefore, the solid sphere has greater acceleration than the solid disk and reaches the bottom of the incline first, regardless of mass differences, as long as rolling without slipping occurs.

Verification / Alternative check:
An alternative way to see this is through energy distribution. Both objects start with the same potential energy if their centres start at the same height. At the bottom, this potential energy is converted to translational kinetic energy (1 / 2 * m * v^2) plus rotational kinetic energy (1 / 2 * I * omega^2). The object with smaller moment of inertia uses less energy in rotation for the same angular speed and can therefore have more translational kinetic energy, meaning higher linear speed at the bottom. Because the sphere has a smaller moment of inertia factor than the disk, it attains a higher linear speed and reaches the bottom earlier.

Why Other Options Are Wrong:
Saying it depends on their masses is incorrect because the acceleration for rolling objects on an incline, under the stated assumptions, is independent of mass and depends only on shape through I / (m * R^2). The solid disk does not reach first; its larger relative moment of inertia leads to smaller acceleration compared to the sphere. Both reaching at the same time would be true only if their effective acceleration down the incline were identical, which is not the case for these different shapes.

Common Pitfalls:
Many students assume that if the bodies have different masses, the heavier one will roll faster, which is not correct for pure rolling without slipping in this idealised scenario. Others may believe that identical heights and radii must lead to identical times. The key point is that distribution of mass (moment of inertia) affects rotational dynamics. Remembering the specific values of I for common shapes and the form a = g * sin(theta) / (1 + I / (m * R^2)) helps avoid such misconceptions.

Final Answer:
Under the given conditions, the object that reaches the bottom first is the solid sphere.