The figure below ( Figure 1 ) illustrates an Atwood\'s machine The figure below
ID: 2258691 • Letter: T
Question
The figure below (Figure 1) illustrates an Atwood's machine
The figure below (Figure 1) illustrates an Atwood's machine Let the masses of blocks A and B be 4.00kg and 2.00kg , respectively, the moment of inertia of the wheel about its axis be 0.300kg?m 2 , and the radius of the wheel be 0.110m . Find the linear accelerations of block A if there is no slipping between the cord and the surface of the wheel. Find the linear accelerations of block B if there is no slipping between the cord and the surface of the wheel Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel. Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel. Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.Explanation / Answer
assume T1 the tension in the rope connected to mass 1 (mass A); T2 is the tension in the other rope
newton's second laws for the masses are
T1 - m 1 g = - m1a (the negative sign occurs because m1, the heavier mass, will descend and down
is the negative direction)
T2 - m2 g = + m2 a
for the wheel, it experiences a net torque of (T1-T2)R because the tensions are different on its two sides
this new torque causes an angular acceleration given by
(T1-T2)R = I alpha where I is themoment of inertia of the disk and alpha the angular acceleration
but, for a wheel, I = 1/2 MR^2 and angular accel = a/R where R is the radius of the wheel and M its mass
therefore (T1-T2)R = 1/2 MR^2 *(a/R) => T1-T2 = 1/2 Ma
subtract the first two equations and get
T1 - T2 - m1g + m2g = - m1 a - m2 a
group and rearrange: a = (m1-m2)g/[1/2 M + m1 +m2]
you are given m1 and m2, find M knowing I = 0.3 = 1/2 MR^2 and R = 0.11m
M=49.58
substitute into the equation and solve for a
a= (4-2)g/[1/2* 49.58 + 4+2]
a=0.63
angular accel = a/R
=0.63/0.11= 5.8
find the tensions by using this value of a in the first two equations for the masses
(T1-m1g = - m1a and T2- m2g = m2a)
T1-m1g = - m1a
T1= m1(g-a)
T1= 4(9.81-0.63)= 36.72N
for T2=
T2- m2g = m2a
T2=m2(g+a)
T2=2(9.81+0.63)
T2=20.88N
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