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Two Masses on a Massed Pulley and a variation

  1. Nov 5, 2012 #1
    1a. The problem statement, all variables and given/known data
    Hi all, I'm having quite a bit of problem on a physics question. I've seen some examples of this problem, but when I follow the suggestions and equations out, the answer I get back is wrong...

    Anyways, here's the problem:
    Given two masses ([itex]m_1,~m_2[/itex]) hanging from opposite sides of a solid disk-shaped pulley of known mass and radius ([itex]m_3,~R[/itex]) from a massless string, determine linear acceleration, angular acceleration of the pulley, the tension between the pulley and each of the two hanging masses and a few other things that I think I can work out on my own. I've attached a picture for reference.

    I'd like to figure out how to solve this generally, but if it helps here are the masses and radii:[tex] m_1 = 3.6\text{ kg}[/tex] [tex] m_2 = 1.7 \text{ kg}[/tex] [tex] m_3 =2.2 \text{ kg} [/tex] [tex] R = 0.15 \text{ m}[/tex]

    1b.
    The variation replaces one of the masses with a sphere rolling without sliding on a flat surface. I've included a picture of this problem as reference. For this problem we are given the mass and radius of the sphere ([itex]m_s,~r_s[/itex]), the mass and radius of the solid disk-shaped pulley ([itex]m_d,~r_d[/itex]), and the mass and the radius of the hoop (which I have been treating as just a point mass because it appears to have no rotation) ([itex] m_h,~r_h[/itex]).

    Again, I'd like to figure this out generally, but here are the values given: $$m_s=3.2\text{ kg},~r_s=0.2\text{ m}$$ $$m_d=1.8\text{ kg},~r_d=0.09\text{ m}$$ $$m_h=2.4\text{ kg},~r_h=0.15\text{ m}$$

    2. Relevant equations
    The equations I'm working with for 1a are: $$\vec{\tau}=\vec{R} \times \vec{F}= RT = I \alpha $$ $$a=R \alpha$$ $$I_\text{disk}= \frac{1}{2}mR^2$$
    For 1b I'm also using: $$I_s= \frac{2}{5}mR^2$$


    3. The attempt at a solution
    Assuming downward acceleration to be negative:
    ##F_\text{net},_1 = m_1 a = T_1 - m_1 g ## resulting in a negative acceleration for ##m_1## ​
    ##F_\text{net},_2 = m_2 a = T_2 - m_2 g ## resulting in a positive acceleration for ##m_2## ​
    Solving both for tension, I get: $$T_1=m_1a+m_1g$$ and $$T_2=m_2a+m_2g$$
    Additionally I have: $$\tau_\text{net}=I \alpha = RT_\text{net} = RT_1 - RT_2$$
    Substituting in ##T_1 ,~ T_2,~\alpha=\frac{a}{R},~I_\text{disk}=\frac{1}{2}m_3 R^2 ##: $$\frac{1}{2}m_3 R^2 (\frac{a}{R}) = Rm_1 a + Rm_1 g - Rm_2 g - Rm_2 a$$
    Simplifying out: $$a = 2g \frac{m_1 - m_2}{2m_2+m_3 - 2m_1}$$
    and this is where I'm stuck. When I plug the numbers in, the online answer program is saying I'm wrong. I also simplified out using ##m_2 a=m_2g -T_2## resulting in: $$a = 2g \frac{m_1 - m_2}{m_3-2m_2 - 2m_1}$$ and ##T_1=m_1g-m_1a## resulting in: $$a = 2g \frac{m_1 - m_2}{m_3-2m_2 + 2m_1}$$ and these were both wrong

    For 1b, I did essentially the same thing, except making down positive: $$\tau_s=I_s\alpha$$ $$\tau_h=R_dT_h$$ $$T_h=m_ha+m_h$$ $$\tau_\text{net}=\tau_s-\tau_h$$ Simplifying and substituting, I get: $$a=\frac{r_dm_hg}{\frac{2}{5}m_sr_s-r_d(m_h-\frac{m_d}{2})}$$ which is wrong.

    Help would be appreciated!
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    These two accelerations mean different things. Either give them separate names or (easier) change the sign of one so that they refer to the same thing. It has led you to get the wrong sign on a term in your answer. I think the rest of the working is OK.
     
  4. Nov 6, 2012 #3
    Herp derp, I can't add. Just reworked it using the right signs and it ended up right. Thanks.
    Ended up with $$a=2g\frac{m_1-m_2}{2m_1+2m_2+m_3}$$
    I'll try doing the right math with the variation problem.
     
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