A disc at rest without slipping, rolls down a hill of height (3 x 9.8) m. What is its speed in m/sec when it reaches the bottom?

Explanation

The problem involves applying the conservation of energy principle. Initially, the disc is at rest at the top of the hill, possessing only potential energy (PE = mgh). As it rolls down, this energy is converted into kinetic energy, which includes both translational (1/2 mv²) and rotational (1/2 Iω²) components. For a disc, the moment of inertia I = 1/2 mR², and the relationship between linear velocity v and angular velocity ω is ω = v/R. Substituting these into the energy equation gives mgh = 1/2 mv² + 1/4 mv², simplifying to mgh = 3/4 mv². Solving for v gives v = sqrt(4gh/3), which evaluates to 19.6 m/s when h is substituted as 3 × 9.8 m.Options A, C, and D are incorrect due to miscalculations or misunderstandings of how energy is distributed and converted during the disc's motion.