A yo-yo has a rotational inertia of 810 g-cm2 and a mass of 180 g. Its axle radi

Question

A yo-yo has a rotational inertia of 810 g-cm2 and a mass of 180 g. Its axle radius is 3.7 mm, and its string is 60 cm long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? 4.7 X cm/s2 (b) How long does it take to reach the end of the string? (c) As it reaches the end of the string, what is its linear speed? cm/s (d) As it reaches the end of the string, what is its translational kinetic energy? (e) As it reaches the end of the string, what is its rotational kinetic energy? (f) As it reaches the end of the string, what is its angular speed? rad/s

Explanation / Answer

Given,

I = 810 g-cm^2 ; m = 180 g ; r = 3.7 mm ; L = 60 cm

a)the angular acceleration will be given by:

aplha = g/R (1/ (1 + I/mR^2))

alpha = 981 cm/s^2/0.37 [1/(1 + 810/180 x 0.37^2) ] = 78.28 m/s^2

a = r alpha

a = 0.37 x 78.28 = 28.96 cm/s^2

Hence, a = 28.96 cm/s^2

b)t = sqrt (2L/a)

t = sqrt (2 x 60/28.96) = 2.036 s

Hence, t = 2.036 s

c)v = a t

v = 28.96 x 2.036 = 58.963 rad/s

Hence, v = 58.963 cm/s

d)E = 1/2 m v^2

E = 0.5 x 0.18 x 0.589^2 = 0.0312 J

Hence, KE = 0.0312 J

[first four parts as per the guidelines has been solved, kindly check and let me know for any query. Rate if this helped you]

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