- #1
dumbperson
- 77
- 0
Homework Statement
I'm trying to find the acceleration of the masses in an atwood machine, the pulley has a rotational inertia I. The pulley has a radius R
Picture : http://en.wikipedia.org/wiki/File:Atwood.svg
But I made it accelerate the other way, so the equations from Newtons 2nd law are (a bit) different for this picture.
Homework Equations
[tex] m_1 \cdot a = T_1 - m_1 \cdot g[/tex]
[tex] m_2 \cdot a = m_2 \cdot g - T_2[/tex]
[tex]Ʃ \tau = \frac{dL}{dt}[/tex]
The Attempt at a Solution
I know you can do it another way, but I wan't to do this with angular momentum L(with respect to midpoint of the pulley). The pulley has a radius R.
[tex] L= I_{pulley} \cdot \omega + (m_1 + m_2) \cdot R^2 \cdot \omega = I_{pulley} \cdot \frac{v}{R} + (m_1 +m_2)vR[/tex]
So:
[tex] \frac{dL}{dt} = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R[/tex]
[tex] Ʃ \tau = T_2 \cdot R - T_1 \cdot R[/tex]
[tex] T_2 = m_2 \cdot g -m_2 \cdot a[/tex]
[tex] T_1 = m_1 \cdot a + m_2 \cdot g [/tex]
So
[tex] Ʃ \tau = g \cdot R (m_2 - m_1) - a(m_2 +m_1)[/tex]
[tex] Ʃ \tau = \frac{dL}{dt}[/tex]
So
[tex] g \cdot R (m_2 - m_1) - a(m_2 +m_1)R = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R [/tex]
Solve for a and i'll get:
[tex] a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+2(m_1+m_2)}[/tex]
The right answer should be:
[tex] a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+m_1+m_2}[/tex]
I notice that if i say:
[tex] Ʃ \tau = gR(m_2-m_1) [/tex] , I do get the right answer. But this is not true is it? this would only be true if the masses are not accelerating? is the angular momentum wrong ?
Where do I go wrong? thank you!
Last edited: