Question

A yo-yo has a rotational inertia of 1100 g·cm² and a mass of 118 g. Its axle radius is 1.99 mm, and its string is 133 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

given :

I = moment of inertia = 1100x10-7 kgm2.

m = mass = 0.118 kg

r = radius = 0.00199 m

L = string length = 1.33 m

see the diagram :

La T mg

For translational motion :

mg - T = ma ------(i)

for rotational motion :

$RT = I\alpha$ ------(ii)

for no slipping at string yo-yo contact :

$a=\alpha R$-----(iii)

a) From (ii) and (iii) :

-----(iv)

a T R2

from (i) and (iv) :

$mg - \frac{Ia}{R^{2}} = ma$

$\Rightarrow a = \frac{mgR^{2}}{I+mR^{2}} = \frac{0.118*9.81*0.00199^{2}}{(1100*10^{-7})+(0.118*0.00199^{2})} = 0.041m/s^{2}$ [answer]

b) applying kinematic equation :

$s = ut +\frac{1}{2}at^{2}$

$\Rightarrow 1.33 = 0+\frac{1}{2}*0.041*t^{2}$

=> t = 8.05 s [answer]

c) applying kinematic equation :

v = u + at

=> v = 0 + 0.041 x 8.05 =  0.33 m/s [answer]

d) Translational kinetic energy = $\frac{1}{2}mv^{2} = \frac{1}{2}*0.118*0.33^2 = 0.0064J$ [answer]

f) as $a=\alpha R$ , we can write

$v=\omega R$

=> $\omega = \frac{v}{R} = \frac{0.33}{0.00199} = 165.83 rad/s$ [answer]

e) Rotational KInetic Energy = $\frac{1}{2}I\omega^{2} = \frac{1}{2}*1100*10^{-7}*165.83^{2}=1.51J$ [answer]