Blocks of mass m1 and m2 are connected by a massless string that passes over the
ID: 1973188 • Letter: B
Question
Blocks of mass m1 and m2 are connected by a massless string that passes over the pulley. The pulley turns on frictionless bearings. Mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest.
a) Assume the pulley is massless. Find the acceleration of m1 and the tension in the string.
b) Suppose the pulley has mass mp and radius R. Find the acceleration of m1 and the tensions in the upper and lower portions of the string. Verify that your answers agree with part a if you set mp = 0.
Explanation / Answer
a)
The net force accelerating m2 is = m2*g - T (with T = tension)
m2g - T = m2a --> T = m2g - m2a.
And the tension is also the force that accelerates m1:
T = m1a
set both equations equal to get
m2g - m2a = m1a --> solve for a:
a(m1 + m2) = m2g
a = m2g/(m1+m2)
T = m1*m2*g/(m1+m2)
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b)
T1 is the tension between m1 and the pulley
T2 is the tension between m2 and the pulley
the net force on m1 is T1 = m1a
the net force on m2 is m2g - T2 = m2a --> T2 = m2g - m2a
the net force on the pulley is T2 - T1
and the torque on the pulley is then = F*R = (T2 - T1)R
and the torque on the pulley is also = I* = I*a/R (with I = moment of inertia of pulley, = angular acceleration of pulley)
--> (T2-T1)R = I*a/R --> divide by R
(T2 - T1) = Ia/R^2 plug in the expressions for T1 and T2:
(m2g - m2a) - m1a = Ia/R^2 --> solve for a:
a(-m2 - m1 - I/R^2) = - m2g
a = m2g/(m1+m2+I/R^2) with I = 1/2*mp*R^2
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plug a into the two equations for T1 and T2:
T1 = m1a
T2 = m2g-m2a