SHOW WORK! The string in a yo-yo is wound around an axle of radius 0.524 cm. The

Question

SHOW WORK!

The string in a yo-yo is wound around an axle of radius 0.524 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.137 kg and outer radius 2.00 cm. Starting from rest, it rotates and falls a distance of 1.14 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. What is the speed of the yo-yo when it reaches the distance of 1.14 m? m/s 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

decrease in potential energy = increase in kinetic energy of system

decrease in PE = 0.137 * 9.81 * 1.14

Increase in KE = 0.5 * 0.137 * V^2 + 0.5 I W^2

V= R1W R1=0.524cm

I = MR2^2/2

from first equation

9.81*1.14 = 0.5V^2 + 0.5*0.5 (R2/R1)^2 V^2

V = 1.6431 m/sec

now for linear motion

mg - T = ma

for rotation

T 0.524 = I alpha

alpha * 0.524 = a

from above equations

g = a + 0.5*(R2/R1)^2a

a = 1.184m/sec^2

V = at

t = 1.387sec

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