Question

problem 8: a yo - yo is rolled up and released from a height of 60 cm. the yo - yo unwinds and reaches the bottom at what speed? assume that the yo - yo can be treated as a uniform disk.

Explanation:

Step1: Conservation of mechanical energy

The initial potential energy \(U = mgh\) (where \(h = 60\space cm=0.6\space m\)) is converted into translational kinetic energy \(K_t=\frac{1}{2}mv^{2}\) and rotational kinetic energy \(K_r=\frac{1}{2}I\omega^{2}\). For a uniform disk, \(I = \frac{1}{2}mr^{2}\), and for rolling without slipping \(\omega=\frac{v}{r}\).

Step2: Substitute \(I\) and \(\omega\) into energy equation

$$
LATEXBLOCK0
$$

Step3: Solve for \(v\)

Cancel out \(m\) from both sides of the equation \(mgh=\frac{3}{4}mv^{2}\), we get \(gh=\frac{3}{4}v^{2}\). Then \(v^{2}=\frac{4gh}{3}\). Substitute \(g = 9.8\space m/s^{2}\) and \(h=0.6\space m\)

$$v=\sqrt{\frac{4\times9.8\times0.6}{3}}$$

$$v=\sqrt{\frac{23.52}{3}}=\sqrt{7.84} = 2.8\space m/s$$

Answer:

\(2.8\space m/s\)