Question

(a) A simple Atwood's machine consists of two masses connected by a string that passes over a pulley. Derive the formula for the acceleration of the masses for general m₁ and m₂ and evaluate for the case m₁ = 3.50 kg and m₂ = 5.50 kg. (b) What assumptions or modifications need to be included if the rotation of the pulley mass M, Radius R is taken into account? If M = 1.00 kg, R = 13.00 cm, determine a value for the modified acceleration a. (c) What percent difference occurs when pulley is ignored?

Answer 1

Take clockwise to be positive for the pulley.

Mg - T₁ = Ma assume M is one block maas and m is another and m_p is mass of pulley

T₂ - mg = ma

rT₁ - rT₂ = ½ m_pr²a

now solve three equation eliminate T₁ and T₂  

we get  T₁ - T₂ = ½ m_p a

Combining the three equations to eliminate the two tensions gives:

(Mg - Ma) - (mg + ma) = ½ m_p a

Mg - mg = Ma + ma + ½ m_p a

a

=

g (M - m) M + m + ½ mp

1. when m_p =0 then a=(5.5-3.5)*9.81/(5.5+3.5) =2.18 m/s²

2. when m_p = 1 kg then a=(5.5-3.5)*9.81/(5.5+3.5+.5*1) =2.065m/s²  

3. percentage differences =(2.18-2.065)*100 / 2.18 =5.275