Question

An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over a pulley. Let the masses be M1 and M2 and M2 = 2M1. Initially, M1 is held fixed a distance y below M2. Find the speed of the blocks when they are the same elevation (that is, the same horizontal position, by then each block has moved y/2).

Answer 1

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

    Em₀ = U₁ + U₂

    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    $E_{mf}$ = K₁ + U₁ + K₂ + U₂

    $E_{mf}$ = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g $y_{f}$ + m₂ g $y_{f}$

Since the masses are joined by a rope, they must have the same speed

     $E_{mf}$ = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g $y_{f}$

   $E_{mf}$= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

How energy is conserved

   Em₀ =  $E_{mf}$

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)