Question

A yo-yo has a rotational inertia of 1250 g·cm2 and a mass of 137 g. Its axle radius is 1.61 mm, and its string is 91.2 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? As it reaches the end of the string, what are its

(c) linear speed,

(d) translational kinetic energy,

(e) rotational kinetic energy,

and (f) angular speed?

Answer 1

From the free body diagram of a yo-yo, using newtons's second law

F_net = ma , we get

mg - T = ma

and , Torque = I? = Ia/r where ? is angular and a is linear acceleration.

so we get

mg = ( m + I/r² ) a

a) solving for a , we get

a = mg / ( m + I/r² )

a = 0.137*9.8 / (0.137 + 0.000125/1.61e-3²)

a = 0.137*9.8 / 48.36

a = 0.0277 m/s²

b) t = sqrt (2h/a)

t = sqrt ( 2*0.912 / 0.0277)

t = 8.115 seconds

c) v = at = 0.2247 m/s

d) K.E = 0.5*m*v²

K.E = 3.458E-3 JOULES

e) R.K.E = 0.5*I*w²

where w = v/r = 139.6

R.K.E = 1.217 joules

f) w = v/r = 139.6 rad/sec

in rev/s , w = 139.6/2*pi

w = 22.22 rev/s