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Grade: 11

                        

Two bodies of masses m1 and m2 are connected by a light string which passes over a frictionless massless pulley .if the pulley is moving upward with uniform acceleration g/2 then tension in the string will be what?

2 years ago

Answers : (1)

Arun
25211 Points
							
Let   m2 > m1.
Consider the masses m1 and m2, with forces Tension T in the string upwards and their weights downwards, acting on them.  Both of these masses are moving with same acceleration a up (m1) or down (m2 as m2>m1), as the string is tight and has a uniform tension all along its length.    This explanation is valid in the frame of reference of moving pulley with an acceleration g/2 upwards.
In the frame of reference of pulley, with an acceleration g/2 upwards, the two masses are having acceleration a and -a.  To balance forces in this non-inertial frame, we need to add a fictitious force of  m1 g/2 downwards on m1 and -m2 g/2 downwards on m2.  Then we write free body diagrams and apply Newtons' laws.
           T - (m1 g +m1 g/2) = m1 a      =>    T - 3 m1*g/2 = m1 a
           (m2 g +m2 g/2) - T = m2 a      =>    3 m2*g/2 - T = m2 a
   Add the two equations to get
                   a = (3g/2) * (m2 - m1) / (m1 + m2)
    then,
                   T = 3m1 g/2 + m1 a  =  3 m1 m2  g  / (m1 + m2)
     Option A is correct.
In the inertial frame wrt  a stationary person, 
   acceleration of the mass m1 = a1 = a + g / 2    upwards
               a1 = g * (2 m2 - m1) / (m1 + m2)      upwards
   acceleration of mass m2 = a2 = a - g / 2        downwards
               a2 =  g * (m2 - 2 m1) / (m1 + m2)    downwards
You can verify,  by putting, m1=m2,  then a = 0 in the frame  of pulley, as both weights are same.  But in the inertial frame, a1 = g/2,  and 
          a2 = -g/2 downwards or,  g/2 upwards.
 
Regards
Arun (askIITians forum expert)
2 years ago
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