Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :
[tex]\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2[/tex]
(c) Moment of inertia :
The net torque acting on it is, [tex]\tau=I\alpha[/tex], I is the moment of inertia
Also, [tex]\tau=Fr[/tex]
So,
[tex]I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2[/tex]
Hence, this is the required solution.
