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Samuelb88

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## Homework Statement

A solid cylinder m2 is attached to an object m1 by a massless rope which passes over a pulley m3 shown below. The cylinder rolls without slipping down the incline and the rope does not slip on the pulley. The system is released from rest when the cylinder is a given distance L from the bottom of the incline.

givens:

m1 = 12. kg

m2 = 48. kg

m3 = 20. kg

r2 = .08 m

r3 = .06 m

I2 = (1/2)*(m2)(.08^2) (moment for cylinder)

I3 = (.4)(m3)(.06^2) (moment for pulley)

Theta = 50. (deg.)

http://img5.imageshack.us/img5/3416/pic3to.jpg

Determine the acceleration of the center of mass of the cylinder.

## Homework Equations

## The Attempt at a Solution

I drew 3 FBDs, one for each object, thus obtaining a system of equations using the 2nd law.

**m1: (object being pulled up)**

Eq. (I):

[tex] m_1a_x = T_1 - m_1g[/tex]

**m3: (pulley)**

rotation is clockwise, thus

[tex]I_3a_z = R_3(T_1 - T_2)[/tex]

Eq. (II):

[tex]= .4m_3a_x = T_1 - T_2[/tex]

**m2: (cylinder)**

m3 is subject to rotational and translational motion, thus

Eq. (III):

[tex]m_2a_x = T_2 + f_s - m_2gsin50[/tex]

Eq. (IV):

[tex]I_2a_z = R_2(T_2 - f_s)[/tex]

[tex]=\frac{m_1a_x}{2}\right) = T_2 - f_s [/tex]

**Adding eq. (III) and (IV):**

[tex](III) + (IV) = \frac{3}{2}\right) m_2a_x = 2T_2 + m_2gsin50[/tex]

^ Eq. (V)

**Adding eq. (I) + (II):**

[tex](I) + (II) = m_1a_x - .4m_3a_x = -T_2 - m_1g[/tex]

^ Eq. (VI)

And adding eq. (V) & (VI):

[tex](V) + (VI) = \frac{3}{2}\right) m_2a_x + 2m_1a_x - .8m_3a_x = m_2gsin50 - 2m_1g[/tex]

Solving for a_x

[tex]a_x = \frac{m_2gsin50 - 2m_1g}{1.5m_2 + 2m_1 - .8m_3}\right) [/tex]

Since the massless rope is attached at the top of the cylinder, the acceleration at the cylinders center of mass is going to be 1/2 of the acceleration shown above.

[tex]a_c_m = \frac{125.275}{160}\right) = .783 m/s^2[/tex]

My professor gives partial answer keys, and the answer is a_cm = .825 m/s^2.

Any help would be greatly appreciated. I'm not sure what i am doing incorrectly here.

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