An Atwood\'s machine consists of blocks of masses m1 = 40.0 kg and m2 = 60.0 kg.

Question

An Atwood's machine consists of blocks of masses m1 = 40.0 kg and m2 = 60.0 kg. Attached by a cord running over a pulley as in the figure. the pulley is a solid cylinder (I=0.5 MR^2) with mass M=20 kg and radius r=0.8m. The block of mass m2 is allowed to drop and the cord turns the pulley without slipping. Assuming the pulley axis is frictionless.

a) Write simplified equations (should be three) with all known parameters substituted to find tension of the string of the left hand side (T1), tension of the string of the right habnd side (T2) and linear acceleration 'a' of the system. You don't have to solve the above equations. Hint: Define positive direction before you start the problem.

For parts b, c, and d assume 'a' = 2.0m/s^2

b) Find the magnitude of the angular acceleration of the pulley.

c) If the m1 mass initially stays rest close to ground, how far does it rise within 10.0 sec?

d) what is the velocity of it after 10.0 sec?

Explanation / Answer

   The given mass m1 = 40kg

The mass m2 =60Kg

The mass of the solid cylinder M = 20kg

The radius of the cylinder R = 0.8m

(a) When the mass m2 is allowed to fall then the tension in the string by mass m2

provides the rotation of the cylinder. If the tensions are equal then there is no rotation and the system is at rest.

(b) From Newtons' law

      m2a = m2g - T2

then T2 = m2g - m2a

and

    m1a = T1 - m1g

       T1 = m1a + m1g

The torque acting on the cylinder

    ? = I? = (T2 - T1) R

        (0.5)MR^2 (a/R) = (m2g - m2a - m1a -m1g)R

        (0.5)Ma = (m2 - m1)g - (m2 + m1) a

     [ (0.5)M + (m1+m2)] a = (m2 - m1) g

Therefore the acceleration of the system

           a = (m2 - m1) g /  [ (0.5)M + (m1+m2)]

              = (60-40) (9.8) /  [ (0.5)(20) + (60+40)]

               = 1.781818181818182 m/s2

(c) The tension

     T1 = m1a + m1g = (40)[ 1.7818+ 9.8]

         = 463.2727272727273N

and the tension

      T2 = m2g - m2a

          = (60)[ 9.8 -1.7818]

         = 481.092N   

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