The Atwood machine in the figure consists of masses m1 = 1.55 kg and m2 = 1.00 k
ID: 2262619 • Letter: T
Question
The Atwood machine in the figure consists of masses m1 = 1.55 kg and m2 = 1.00 kg and a solid pulley with a diameter of 6.7 cm and a mass mp = 0.526- kg. When released from rest, what is the speed of the heavier mass after it has fallen a distance of 1.65 m? Treat the pulley as a uniform disk. Please show all work.
The Atwood machine in the figure consists of masses m1 = 1.55 kg and m2 = 1.00 kg and a solid pulley with a diameter of 6.7 cm and a mass mp = 0.526- kg. When released from rest, what is the speed of the heavier mass after it has fallen a distance of 1.65 m? Treat the pulley as a uniform disk. Please show all work.Explanation / Answer
use the moment of inertia for the pulley, which consider a disk, so its moment of inertia is given by
I = 1/2 m r ^2 = 0.5 (0.526) (6.7/2 * 10^-2) ^2 = 2.95 *10^(- 4) Kg . m^2
Let T1 be the tension in the cable acting at m1, and T2 be the tension acting on m2
The m1g - T1 = m1 a, and T2 - m2 g = m2 a
Note that the acceleration of m1 is a = r d2 theta /dt2 = r * alpha
The net moment on the pully is
r ( T1 - T2) = r ( m1(g - a ) - m2 (a + g) ) = r ( - a (m1+m2) + g(m1 - m2) )
Therefore
I * a / r = r ( - a(m1 +m2) + g(m1 - m2) )
Therefore, a ( I / r^2 + (m1 + m2) ) = g (m1 - m2)
i.e.
a ( 1/2 (0.526) + (1.55) + 1) ) = 9.8 (1.55 - 1)
so a = 1.916 m/s^2
Applying
v2^2 - v1^2 = 2 a d, then
v2 = sqrt( 2 (1.916) (1.65) ) = 2.51 m/s
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