- #1

PhDeezNutz

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- Homework Statement
- See Pic Below

- Relevant Equations
- ##v_f^2 = v_i^2 + 2a \Delta y##

##I_{ring} = m_w R^2##

## a = R \alpha##

##\sum \tau = I \alpha##

The equation that connects final velocity with distance traveled is

##v_f^2 = v_i^2 + 2a \Delta y##

Since the system starts from rest ##v_i = 0##

and the above equation becomes.

##v_f^2 = 2a \Delta y##

Since there is rotation in this system we need to connect ##a## to the rotation of the wheel. ##a = \alpha R##. So that becomes

##v_f^2 = 2 \alpha R \Delta y##

Now we need to find ##\alpha## my summing the torques and setting it equal to ##I \alpha## where ##I = m_w R^2##. Although R wasn't given in the problem statement I figure we still need it even though it will cancel out in the end.

##\sum \tau = I \alpha##

I'm going to choose the counterclockwise direction as positive.

##MgR - mgR = m_w R^2 \alpha##

##\alpha = \frac{g R\left(M-m \right)}{m_w R^2} = \frac{g \left(M - m \right)}{m_w R}##

So

##v_f^2 = 2 \left(\frac{g\left( M - m\right)}{m_w R} \right) R \Delta y##

##v_f = \sqrt{\frac{2g \left(M-m\right)}{m_w} \Delta y}##

When I plug in ##M = 20##, ##m = 5##, ##m_w = 10##, and ##\Delta y = 1## I get

##v_f = 5.42##

Where did I go wrong? This is different than the stated answer.