Question

In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first

Answer 1

Answer:

solid cylinder

Explanation:

the object that arrives first is the object that has more speed, let's use the concepts of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point

         Em_f = K = ½ mv² + ½ I w²

since the body has rotational and translational movement

how energy is conserved

         m g h = ½ mv² + ½ I w²

linear and angular velocity are related

          v = w r

          w = v / r

we substitute

          m g h = ½ mv² + ½ I (v/r) ²

          mg h = ½ v² (m + I /r²)

          v = $\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }$

           

the tabulated moments of inertia for the bodies are

solid cylinder     I = ½ m r²

hollow cylinder  I = m r²

we look for the speed for each body

solid cylinder

          v₁ = $\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }$

          v₁ = $\sqrt{2gh} \ \sqrt{2/3}$

let's call    v₀ = $\sqrt{2gh}$

         v₁ = 0.816 v₀

hollow cylinder

          v₂ = $\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }$

          v₂ = v₀ √½

          v₂ = 0.707 v₀

Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive