The string in a yo-yo is wound around an axle of radius 0.530 cm. The yo-yo has

Question

The string in a yo-yo is wound around an axle of radius 0.530 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.351 kg and outer radius 1.95 cm. Starting from rest, it rotates and falls a distance of 1.18 m (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of 1.18 m? m/s (b) How long does it take to fall? [Hint: The translational and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.] s

Explanation / Answer

Height energy lost = Kinetic energy gained

KE gained = translational KE + Rotational KE

KE = 0.5 * 0.351 * V^2 + 0.5 * (0.5 * 0.351 *0.0195^2) * (V/ 0.00530)^2

KE = 0.1755*V^2 + 1.187 V^2 = 1.3625 . V^2 J

delta GPE = m *g *delta h = 0.351 * 9.8 * 1.18 = 4.06 J

4.06 = 1.3625V^2

V = 1.72 m/s

2) Time taken = distance fallen / average velocity

T = 1.38 / 0.86 = 1.60 s

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