Final answer:
The final velocity of a hoop rolling without slipping down a 5.10-meter-high hill is approximately 10.0 m/s, calculated using the conservation of mechanical energy.
Step-by-step explanation:
To determine the final velocity of a hoop that rolls without slipping down a 5.10 m high hill, we can employ the conservation of mechanical energy principle, assuming no energy is lost to friction. The potential energy (PE) of the hoop at the top of the hill will convert into both translational kinetic energy (KEtrans) and rotational kinetic energy (KErot) at the bottom.
At the top, PE = mgh, where m is the mass of the hoop, g is the acceleration due to gravity, and h is the height. At the bottom, KEtrans = (1/2)mv2 and KErot = (1/2)I2.
For a hoop, I = mr2 and v = r, so KErot = (1/2)mv2, the same as KEtrans. Therefore, mgh = mv2 (since there are two components of kinetic energy, this equation is multiplied by 2).
The mass m cancels out, leading to gh = v2. Solving for v, the final velocity v = sqrt(2gh). Substituting the given height of 5.10 m and g as 9.81 m/s2, we get:
v = sqrt(2 * 9.81 m/s2 * 5.10 m)
v ≈ sqrt(100.062) m/s ≈ 10.0 m/s.