Question

what is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.00 - m - high hill, starting from rest?
enter a number.
m/s

Explanation:

Step1: Apply conservation of mechanical energy

The initial energy \(E_{i}\) is gravitational potential energy \(E_{i}=mgh\) (since it starts from rest, \(K_{i} = 0\)). The final energy \(E_{f}\) is the sum of translational kinetic energy \(K_{t}=\frac{1}{2}mv^{2}\) and rotational kinetic energy \(K_{r}=\frac{1}{2}I\omega^{2}\). For a hoop, \(I = mr^{2}\), and since it rolls without - slipping \(\omega=\frac{v}{r}\). Then \(E_{f}=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v}{r})^{2}=mv^{2}\)

Step2: Set \(E_{i}=E_{f}\)

$$mgh=mv^{2}$$

$$v = \sqrt{gh}$$

Substitute \(g = 9.8\space m/s^{2}\) and \(h=3.00\space m\)

$$v=\sqrt{9.8\times3.00}=\sqrt{29.4}\approx5.42\space m/s$$

Answer:

\(5.42\)