The string in a yo-yo is wound around an axle of radius 0.537 cm. The yo-yo has

Question

The string in a yo-yo is wound around an axle of radius 0.537 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.219 kg and outer radius 1.95 cm. Starting from rest, it rotates and falls a distance of 1.03 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 1.03 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.]

Explanation / Answer

Q ) The string in a yo-yo is wound around an axle of radius 0.505 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.182 kg and outer radius 2.25 cm. Starting from rest, it rotates and falls a distance of 1.38 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. 1.What is the speed of the yo-yo when it reaches the distance of 1.38 m? 2.How long does it take to fall? ANS ) Height energy lost = Kinetic energy gained KE gained = translational KE + Rotational KE KE = 0.5 . 0.182 . V^2 + 0.5 . (0.5 . 0.182 . 0.0225^2) . (V/ 0.00505)^2 KE = 0.091 .V^2 + 0.903 .V^2 = 0.994 . V^2 J ?GPE = m . g . ?h = 0.182 . 9.8 . 1.38 = 2.46 J 2.46 = 0.994 V^2 V = 1.57 m/s 2) Time taken = distance fallen / average velocity T = 1.38 / 0.787 = 1.75 s ..........................simliar kind of problem

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