The string in a yo-yo is wound around an axle of radius 0.504 cm. The yo-yo has

Question

The string in a yo-yo is wound around an axle of radius 0.504 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.298 kg and outer radius 1.99 cm. Starting from rest, it rotates and falls a distance of 1.22 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. 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.]

Explanation / Answer

Height energy lost = Kinetic energy gained KE gained = translational KE + Rotational KE KE = 0.5*0.298*V^2 + 0.5 *(0.5*0.298*0.0199^2) . (V/ 0.00504)^2 KE = 0.149.V^2 + 1.161.V^2 = 1.310 V^2 J ?GPE = m . g . ?h = 0.298* 9.8 * 1.22 = 3.56 J 3.56 = 1.31V^2 V = 2.72 m/s 2) Time taken = distance fallen / average velocity T = 1.22 /( 2.72/2)= 0.89

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