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**[SOLVED] Pulley problem (Rotation)**

Mass m1 = 9.07 kg sits on a frictionless, horizontal surface. A massless string is tied to m1, passes over a pulley (a solid disk of mass Mp = 5.42 kg and radius Rp = 37.9 cm), and is tied to m2 = 6.94 kg hanging in space.

Find the acceleration of the masses.

It seems I'm having some trouble with this problem. I'm not sure if my algebra is right or if I am doing the wrong thing with the moment of Inertia. I was hoping someone could help me.

torque = I [tex]\alpha[/tex]

T = ma

a = R[tex]\alpha[/tex]

My attempt is as follows:

The question assumes that there are non-slip conditions and it starts from rest.

1 . First, I find the pulley's torque

T1 = ma

T2 = mg - ma

torque = I [tex]\alpha[/tex]

r T = I [tex]\alpha[/tex]

T2 R - T1 R = I [tex]\alpha[/tex]

2. Second, I find the acceleration.

a = R [tex]\alpha[/tex]

a / R = [tex]\alpha[/tex]

3. Third, I solve for a.

T2 R - T1 R = I[tex]\alpha[/tex]

R (T2 - T1) = I (a/R)

T2 - T1 = I (a/R) / R

T2 - T1 = I a / R[tex]^{2}[/tex]

m2g-m2a - m1a = I a/R[tex]^{2}[/tex]

m2g = m2a + m1a + I a/R[tex]^{2}[/tex]

m2g = a (m2 + m1 + I /R[tex]^{2}[/tex]

m2g / m2 + m1 + I / R[tex]^{2}[/tex] = a

(6.49)(9.81) / (I/R[tex]^{2}[/tex]) + 9.07 +6.94 = a

I = mass of pulley * radius[tex]^{2}[/tex]

I = (5.42 kg)(.379 m)[tex]^{2}[/tex]

I = .7785

(6.49)(9.81) / (.7785/.379[tex]^{2}[/tex]) + 9.07+6.94 = a

63.667 / 5.420 + 9.07 + 6.94 =

63.667/21.43 = 2.971 m/s[tex]^{2}[/tex]

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