Rolling and translational dynamics problem

[Samuelb88]

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.

 

[anathema]

Thank you for sharing your solution for this problem. It seems like you have a good understanding of the concepts and have approached the problem correctly. However, there are a few things that could potentially explain the discrepancy between your answer and the answer given by your professor.

1. Significant figures: Your solution includes a calculation where you divide by 160, which has two significant figures. This means that your final answer should also have two significant figures, which would make it 0.78 m/s^2. However, your professor's answer has three significant figures, which could suggest that they were using more precise values for the given variables.

2. Trigonometry: In your solution, you use the value of 50 degrees for the angle theta. However, in the diagram provided, the angle seems to be closer to 40 degrees. This could have a small impact on the final answer.

3. Rounding errors: In your solution, you have rounded some numbers to two decimal places, which could introduce small errors in your final answer. It is always best to keep as many decimal places as possible in intermediate calculations to minimize rounding errors.

I hope this helps to clarify any potential issues with your solution. Keep up the good work and keep practicing! Science is all about trial and error, and it's important to keep trying and learning from mistakes. Good luck!

 

[kerimek]

Hello,

Thank you for your detailed attempt at solving this problem. Your approach seems reasonable and your equations look correct. However, there are a few things that may be causing the discrepancy between your answer and the given answer.

Firstly, it seems like you may have made a mistake in your calculation of the moment of inertia for the pulley. The correct moment of inertia for a pulley is I = (1/2)mr^2, not (0.4)mr^2. This may change the value of a_cm slightly.

Secondly, it is important to note that the given answer may be rounded, so there could be a small margin of error in your answer. Additionally, the given answer may have been calculated using a different method or approach, resulting in a slightly different answer.

Finally, it is always a good idea to double check your calculations and make sure you are using the correct units and values for all variables. Sometimes, small errors in calculations or unit conversions can lead to significant differences in the final answer.

Overall, your approach and solution seem correct and it is possible that the discrepancy in the final answer is due to a small calculation error or rounding. I hope this helps and good luck with your studies!