2 Masses and a Wheel (with mass)

• PhDeezNutz
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.

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.

PhDeezNutz
PS I don't get any of those answers!

PhDeezNutz
PeroK said:
PS I don't get any of those answers!
Did you get 2.89 m/s?

PhDeezNutz said:
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:
$$a = \frac{M-m}{M+m+m_w}g$$

PhDeezNutz
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
PeroK said:
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##

## 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}##

PeroK

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