Math Problem Statement

Two uniform solid cylinders of the same size but different mass are released simultaneously at the top of an incline and roll without slipping to the bottom of the incline. Which of the following statements is true?

Solution

This problem involves two solid cylinders of identical size but different masses rolling down an incline. The key concepts here are rotational dynamics and energy conservation.

Since the cylinders roll without slipping, we can consider both translational and rotational kinetic energy at the bottom. Let’s analyze the motion using the concept of energy conservation and note that the moment of inertia for a solid cylinder is I=12mr2I = \frac{1}{2} m r^2.

  1. Energy Conservation:

    • Initial potential energy at the top of the incline is converted to kinetic energy (both translational and rotational) at the bottom.
    • The initial gravitational potential energy for each cylinder is mghmgh, where mm is the mass, gg is acceleration due to gravity, and hh is the height of the incline.
  2. Kinetic Energy at the Bottom:

    • When rolling without slipping, total kinetic energy KEKE at the bottom will be the sum of translational and rotational kinetic energy: KE=12mv2+12Iω2KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
    • Since v=rωv = r\omega, we can substitute I=12mr2I = \frac{1}{2}mr^2 and simplify, leading to: KE=12mv2+14mv2=34mv2KE = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2
  3. Solving for Final Speed:

    • Using conservation of energy, mgh=34mv2mgh = \frac{3}{4}mv^2.
    • Solving for vv, we get v=4gh3v = \sqrt{\frac{4gh}{3}}, which is independent of mass.
  4. Conclusion:

    • Since both cylinders have the same size and shape, they will reach the bottom simultaneously, regardless of their masses.

Therefore, the correct answer is: (b) Both cylinders reach the bottom of the incline at the same time.

Would you like a deeper explanation of any part?

Here are some related questions:

  1. What would happen if the cylinders were different in shape (e.g., one cylinder and one sphere)?
  2. How would the answer change if the incline was frictionless and the cylinders could not roll?
  3. How does the distribution of mass affect rotational motion in rolling objects?
  4. If the cylinders had different radii but the same mass, would they still reach the bottom at the same time?
  5. How does the rotational inertia influence the speed of rolling objects on an incline?

Tip: When rolling without slipping, both rotational and translational kinetic energy should be considered to apply conservation of energy effectively.

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Math Problem Analysis

Mathematical Concepts

Physics
Rotational Dynamics
Energy Conservation

Formulas

Moment of inertia for a solid cylinder: I = (1/2)mr^2
Total kinetic energy: KE = (1/2)mv^2 + (1/2)Iω^2
Conservation of energy: mgh = (3/4)mv^2

Theorems

Conservation of Energy

Suitable Grade Level

Undergraduate Physics