Unequal masses M and m are suspended from a pulley as shown in the gure. (a) Ana

Question

Unequal masses M and m are suspended from a pulley as shown in the gure. (a) Analyze the forces in which mass m participates, using a table in the format shown in section 5.3. [The forces in which the other mass participates will of course be similar, but not numerically the same.] (b) Find the magnitude of the accelerations of the two masses. [Hints: (1) Pick a coordinate system, and use positive and nega- tive signs consistently to indicate the directions of the forces and accelerations. (2) The two accelerations of the two masses have to be equal in magnitude but of opposite signs, since one side eats up rope at the same rate at which the other side pays it out. (3) You need to apply Newton's second law twice, once to each mass, and then solve the two equations for the unknowns: the acceleration, a , and the tension in the rope, T .] p (c) Many people expect that in the special case of M = m , the two masses will naturally settle down to an equilibrium position side by side. Based on your answer from part b, is this correct? (d) Find the tension in the rope, T . p (e) Interpret your equation from part d in the special case where one of the masses is zero. Here interpret" means to figure out what hap- pens mathematically, figure out what should happen physically, and connect the two.

Explanation / Answer

From a free body diagram about m1 you get T1 - m1g = m1a or T1 = m1g + m1a

From the second body we have m2g - T2 = m2a or T2 = m2g - m2a

About the pullet when we sum torques we have T2*r - T1*r - 0.25 = m*r^2*alpha ( I'm assuming the pulley can be modeled as a thin walled cylinder so I = m*r^2) Now alpha = a/r. So
this eqn becomes T2r - T1r -0.25 = m*r*a

Now substitute T1 and T2 into this eqn

(m2g - m2a)*r - (m1g + m1a)*r -0.25 = m*r*a

Solving for a we get a (m1 + m2 + m) = g*(m2 - m1) - 0.25/r

So a = [g*(m2 - m1) -0.25/r]/9m1 + m2 + m) = [9.8*(0.85 - 0.40) -0.25/0.25]/(0.85 + 0.40 +0.30) = 2.20 m/s^2

Now since 2.20m/s^2 is not the answer, it is possible that the pulley is to be modeled as a solid disk. Then I = 1/2*m*r^2 not m*r^2

If that's the case the the governing eqn from above
((m2g - m2a)*r - (m1g + m1a)*r -0.25 = m*r*a)

becomes (m2g-m2a)*r -(m1g+m1a)*r -0.25 = 1/2m*r*a

Solving this yields a (m1 + m2 + m/2) = g*(m2 - m1) - 0.25/r

So a = [g*(m2 - m1) -0.25/r]/(m1 + m2 + m/2) = [9.8*(0.85 - 0.40) -0.25/0.25]/(0.85 + 0.40 +0.30/2) = 2.44 m/s^2

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