Respuesta :
Answer:
  a = 7.29 m / s²,    T = 0.40 N
Explanation:
To solve this exercise we must apply Newton's second law to each body
The needle
       W -T = m a
       mg - T = ma
The spool, which we will approach by a cylinder
       Σ τ = I α
       T R = I α
the moment of inertia of a cylinder with an axis through its center is
       I = ½ M R²
angular and linear variables are related
      a = α R
      α = a / R
we substitute
      T R = (½ M R²) a / R
      T = ½ M a
we write our system of equations together
       mg - T = m a
           T = ½ M a
we solve
       m g = (m + ½ M) a
       a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]
let's calculate
       a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]
       a = 7.29 m / s²
now we can look for the tension
       T = ½ M a
       T = ½ 0.110 7.29
        T = 0.40 N
