Question

An Atwood's machine consists of blocks of masses m₁ = 9.1 kg and m₂ = 20.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 8.00 kg and radius r = 0.200 m. The block of mass m₂ is allowed to drop, and the cord turns the pulley without slipping.

a) Why must tension T2 be greater than T1

B) what is the acceleration of the system assuming there is no friction

C) Tension in T1=

Tension in T2=

Answer 1

Given that mass m_1 = 9.1 kg , mass m_2 = 20.0 kg

mass of the solid cyllinder is M = 8.00 kg , radius r = 0.200m

  a ) Since the mass m_2 > m_1 then the tension T_2 is greater than the tens ion T_1

    Apply Newtons law for each mass then

    for mass m_1

                      T_1 - m_1 *g = m_1 a

                                  T_1 = m_1 g + m_1 a   --------------(1)

    for mass m_2

                          m_2 g - T_2   = m_2 a

                                      T_2 = m_2 g - m_2 a     -------------(2)

    From the pully

                              T_2 *r- T_1 *r = I ?

                               T_2 *r- T_1 *r = I a / r

                                       T_2 - T_1 = I a/ r^2
                                                    T_ 2 - T_1 = (M r^2 / 2)a/ r^2

                                        T_2 - T_ 1 = ( M/2 )a   --------------- (3)

                         m_2 g -m_a - m_ 1g - m_ 1a = M a /2

                                    a = ( m_2 - m_1 )g / ( m_1 + m_2 + M /2)      

                                        =- ----------m/ s^2   

               

substitute the a value in the eqation (1) then we getT_1 =---------- N

  

substitute the a value in the eqation (2) then we get T_ 2=---------- N