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Blocks of mass m 1 and m 2 are connected by a massless string that passes over t

ID: 2240126 • Letter: B

Question

Blocks of mass m1 and m2 are connected by a massless string that passes over the pulley in the figure . The pulley turns on frictionless bearings, and mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest.


Part A

Assume the pulley is massless. Find the acceleration of m1.

Express your answer in terms of the given quantities.

ANSWER:

  a1 =


Part B

Find the tension in the string.

Express your answer in terms of the given quantities.

ANSWER:

  T =


Part C

Suppose the pulley has mass mp and radius R. Find the acceleration of m1. Verify that your answers agree with part A if you set mp=0.

Express your answer in terms of the given quantities.

ANSWER:

  a1 =


Part D

Find the tension in the upper portion of the string. Verify that your answers agree with part B if you set mp=0.

Express your answer in terms of the given quantities.

ANSWER:

  Tupper =


Part E

Find the tension in the lower portions of the string. Verify that your answers agree with part B if you set mp=0.

Express your answer in terms of the given quantities.

ANSWER:


  Tlower =


  a1 =


Blocks of mass m1 and m2 are connected by a massless string that passes over the pulley in the figure . The pulley turns on frictionless bearings, and mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest. Assume the pulley is massless. Find the acceleration of m1. Express your answer in terms of the given quantities.

Explanation / Answer


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)
------------------------
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
----
plug a into the two equations for T1 and T2:
T1 = m1a
T2 = m2g-m2a