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Rotational Motion and linear accerlation 
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#1
Dec1809, 01:14 PM

P: 54

1. The problem statement, all variables and given/known data
A solid cylinder of weight 50 lb and radius 3.0 inches has a light thin tape wound around it. The tap passes over a light smooth fixed pulley to a 10 lb body hanging vertically in the air. If the plane on which the cylinder moves is inclined 30 degrees to the horizontal, find (a) the linear acceleration of the cylinder down the inclined plane, and (b) the tension of the tabpe, assuming no slippage. 2. Relevant equations [tex]\sum _\tau = I_c\alpha[/tex] [tex]I_c = \frac{M_c R_0^2}{2}[/tex] 3. The attempt at a solution I'm having an issue seeing the relationship linear acceleration and Newton's 2nd law for rotational motion. Edit: See next post 


#2
Dec1809, 04:10 PM

P: 54

Ok... scratch what I said earlier.
[tex]\alpha = \frac{\tau}{I_c} [/tex] [tex]\alpha = \frac{a_t}{R} [/tex] So.. [tex]\frac{\tau}{I_c} = \frac{a_t}{R}[/tex] which means [tex]\tau R = a_t I_c[/tex] (*) A force diagram tells me [tex]\tau = m_Ag sin\vartheta  m_Bg[/tex] The cylinder rolls down the incline [tex]\vartheta[/tex] under the influence of gravity and the weight generated by the blocks mass detracts from that value via the tape. So, plugging [tex]I_c[/tex] for a solid cylinder and the derived value for [tex]\tau[/tex] into * we get: [tex]m_Ag sin\vartheta  m_Bg = \frac{m_AR^2a_t}{2} [/tex] Meaning that [tex]a_t[/tex] (or the linear accleration) would be: [tex]a_t = 2 (\frac{gsin\vartheta}{R}  \frac{m_Bg}{m_aR^2}) [/tex] Granted.. I didn't take the units into consideration.. but I am hoping for a more conceptual and general understanding of the problem, units aside. Am I closer to being on the right track? 


#3
Dec1809, 04:53 PM

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#4
Dec1809, 05:21 PM

P: 54

Rotational Motion and linear accerlation
I'm calculating [tex]T = m_Bg[/tex] as the m_B is free hanging and pulling on the tape. 


#5
Dec1809, 05:27 PM

P: 54

I've attached a picture of the set up.... just for reference.
Thanks for the assistance. 


#6
Dec1809, 05:30 PM

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#7
Dec1809, 05:36 PM

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#8
Dec1809, 05:39 PM

P: 54

Well... I have had trouble with Torque vs Force.... So I'm guessing [tex]\alpha = \frac{\tau}{I_c}[/tex] isn't valid? If it is.. how to we calculate tau?
Concerning Tension... Would [tex]T_{Net} = m_Agsin\vartheta  m_Bg[/tex] ? 


#9
Dec1809, 05:41 PM

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#10
Dec1809, 06:04 PM

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#11
Dec1809, 06:13 PM

P: 54

I suppose that means that torque is force applied radially. So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape? Also, does the Force in the equation for torque equal the force of the weight of the block pulling through the tape? Idea: Edit:see better idea [tex]\tau = Rm_Agsin\vartheta[/tex] Therefore, [tex]R^2m_Agsin\vartheta = \frac{m_AR^2a_t}{2}[/tex] or [tex]gsin\vartheta = \frac{a_t}{2}[/tex] [tex]a_t = \frac{gsin\vartheta}{2}[/tex] right? closer at least? Better idea: [tex]\tau = r \times F[/tex] [tex]\tau = r m_Agsin\vartheta m_Bg[/tex] actually.. this is right back where I started........... The pulley has negligible mass and is fixed... so wouldn't the tension in either piece of the tape be the same? 


#12
Dec1809, 06:53 PM

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#13
Dec1809, 06:57 PM

P: 54

So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape?



#14
Dec1809, 07:07 PM

P: 54

[tex]\tau = R \times F = Rm_Agsin\vartheta + T[/tex] [tex]F_B = ma = m_Bg T[/tex] .:. [tex]Rm_Agsin\vartheta = m_Bg[/tex] There's no unknown in this equation. What is it that I am missing? Should I be solving for a system? Edit: IDEA! [tex]\tau = R \times F = RTsin\vartheta[/tex] However.... I'm still seeing [tex]F_B = ma = m_Bg T[/tex] which says [tex]T = m_Bg[/tex] which you've told me isn't the case............. 


#15
Dec1809, 07:48 PM

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Do it step by step. 


#16
Dec1809, 08:44 PM

P: 54

Ok... so
Forces for cylinder: Linear [tex]m_Agsin\vartheta  T = m_Aa[/tex] => [tex]T = m_Agsin\vartheta m_Aa[/tex] Radial [tex]\tau = R \times F = R m_Aasin\vartheta = I_c\alpha[/tex] => [tex]\alpha = \frac{R m_Aasing\vartheta}{I_c}[/tex] Forces for block: [tex]T m_Bg = m_Ba[/tex] => [tex]T = m_Ba + m_Bg[/tex] .:. Solving system for T => [tex]m_Agsin\vartheta m_Aa = m_Ba + m_Bg[/tex] So [tex]a = \frac{m_Agsin\vartheta  m_Bg}{m_A + m_B}[/tex] BUT... linear acceleration from angular: [tex]a = R\alpha[/tex] I don't believe that I'm using the right value for Force in the torque equation.... perhaps other problems.... Am I even using the right base equations.... it's starting to seem like maybe that's a no. I do, however, appriciate the ever so gentle guidance (in all sincerity).... I'll have done every possible algebraic operation to these equations before it's over.... I look forward to seeing what angle I should work next. 


#17
Dec1909, 07:36 AM

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P: 41,306

The big problems are: (1) Not recognizing that the two linear accelerations are different. (2) Using the wrong force in your torque equation. 


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