2 Masses and a Wheel (with mass)

[PhDeezNutz]

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.

 

[PeroK]

You can't assume that the mass $M$ creates a tension of $Mg$ in the string. If it did, it would not iself be accelerating.

 

[PeroK]

PS I don't get any of those answers!

 

[PhDeezNutz]

PS I don't get any of those answers!

Did you get 2.89 m/s?

 

[PeroK]

Did you get 2.89 m/s?

Yes, with $g = 9.8m/s^2$.

Here's a quick way. Note that $M, m$ and every mass element on the rim of the wheel must have the same speed $v$. The total external force on the system is $Mg - mg$, so we have:
$$a = \frac{M-m}{M+m+m_w}g$$

 

[PeroK]

PS From the kinematic equation, we have:
$$v^2 = 2as$$And this is also what we get from the energy equation:
$$\frac 1 2(M + m + m_w)v^2 = (M-m)gs$$Where $s$ is the distance $M$ moves.

You can also verify the trick, as the total KE is:
$$KE = \frac 1 2(M+m)v^2 + \frac 1 2 I_w \omega^2$$And$$I_w \omega^2 = m_wR^2(\frac v R)^2 = m_wv^2$$

 

[PhDeezNutz]

Yes, with $g = 9.8m/s^2$.

Here's a quick way. Note that $M, m$ and every mass element on the rim of the wheel must have the same speed $v$. The total external force on the system is $Mg - mg$, so we have:
$$a = \frac{M-m}{M+m+m_w}g$$

That’s exactly what I got!!

Just to hash out details.

Torque Equation on the Wheel

$m_w R^2 \alpha = m_w R^2 \frac{a}{R} = m_w aR = F_1 R - F_2 R$

Force on $M$

$Mg - F_1 = Ma$

$F_2 - mg = ma$

So

$F_1 = M \left( g - a\right)$
$F_2 = m\left( a + g \right)$

Plugging that into the sum of torques equation

$m_w a R = MgR - MaR - maR -mgR$

leading to your result

$a = \frac{\left( M-m \right)g}{m_w + m + M}$

All that is left to do is plug it into $v_f = \sqrt{2 a \Delta y}$