the string in a yo-yo is wound around an axle of radius 0.470 cm. The yo-yo has
ID: 1695073 • Letter: T
Question
the string in a yo-yo is wound around an axle of radius 0.470 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.185 kg and outer radius 1.90 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?
answer is in m/s.
i'm lost on how the translational and rotational energies are related, but the yo-yo is not rolling on its outer radius.
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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