Respuesta :
Answer:
 w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]
Explanation:
For this exercise let's start by applying Newton's second law to the mass with the string
        W - T = m a
In this case, as the system is going down, we will assume the vertical directional down as positive.
        T = W - m a
Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations
        ∑ τ = I α
        T r = I α
the moment of inertia of the disk is
        I = ½ M R²
angular and linear acceleration are related
        a = α r
we substitute
        T r = (½ m R²) (a / r)
        T = ½ m ([tex]\frac{R}{r}[/tex] )² a
we write our two equations
        T = W - m a
        T = ½ m ([tex]\frac{R}{r}[/tex] )² a
we solve the system of equations
       W - m a = ½ m (\frac{R}{r} )² a
       m g = m a [ 1 + ½ (\frac{R}{r} )² ]
       a = [tex]\frac{g}{ 1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]
this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations
       w² = w₀² + 2 α θ
       v² = v₀² + 2 a y
as the system is released its initial angular velocity is zero
       w² = 0 + 2 α θ
       v² = 0 + 2 a y
we look for the angular acceleration
       a =α r
       α = a / r
       α = [tex]\frac{g}{r (1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]
we look for the angle, remember that they must be measured in radians
       θ = s / r
in this case we approximate the arc to the distance
      s = y
      θ = y / r
we substitute
      w = [tex]\sqrt{2 \frac{g}{ r( \frac{1}{2} (\frac{R}{r})^2 } \frac{y}{r} }[/tex]
      w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]
  for the simple case where r = R
      w = [tex]\sqrt{ \frac{2gy}{ \frac{3}{2} R^2 } }[/tex]
      w = [tex]\sqrt{ \frac{4}{3} \frac{gy}{R^2} }[/tex]
