A yo-yo has a rotational inertia of 810 g middot cm^2 and a mass of 140 g. Its a

Question

A yo-yo has a rotational inertia of 810 g middot cm^2 and a mass of 140 g. Its axle radius is 4.5 mm, and its string is 130 cm long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (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? (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?

Explanation / Answer

a)

for the acceleration of the yo yo

linear acceleration = net force/(m + I/r^2)

linear acceleration = 0.140 * 9.8/(0.140 + 0.810 * 0.01^2/.0045^2)

linear acceleration = 0.331 m/s^2

linear acceleration = 33.1 cm/s^2

b)

let the time taken is t

130 = 0.50 * 33.1 * t^2

t = 2.80 s

the time taken is 2.80 s

c)

linear speed = 33.1 * 2.80 cm/s

linear speed = 92.7 cm/s = 0.927 m/s

d)

translational kinetic energy = 0.50 * 0.140 * 0.927^2

translational kinetic energy = 0.0601 J

e)

rotational kinetic energy = 0.50 * I * w^2

rotational kinetic energy = 0.50 * 0.810 * 0.01^2 * (0.927/.0045)^2

rotational kinetic energy = 1.72 J

f)

final angular speed = v/r

final angular speed = 0.927/.0045

final angular speed = 206 rad/s

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