Double Pendulum With Acceleration

[matineesuxxx]

Homework Statement

You are holding a special pendulum with two masses $m_1$ and $m_2$, instead of one, connected by a rope as shown in Fig. You lower the pendulum such that the tension in the rope between the two masses is half the weight of the bottom mass. Find the acceleration with which you lower the pendulum.

Homework Equations

$\sum \hat{\text{F}} = m\hat{a}$

$\text{w} = m\hat{g}$

The Attempt at a Solution

I thought that I simply needed to consider the bottom mass, $m_2$ and so since the forces acting on it are gravity and the tension between the two masses, I set up my equations as such:

$\sum \hat{\text{F}}_\text{y} = \hat{\text{T}}_{1,2} - m_2\hat{g} = -m_2\hat{a}$, where $\hat{\text{T}}_{1,2}=\frac{1}{2}m_2 \hat{g}$

so I end up with $\hat{a} = \frac{\hat{g}}{2}$, however the book gives an answer of

$\hat{a} = \frac{m_2\hat{g}}{m_1 + m_2}$ but I can only get that answer if I start with

$\sum \hat{\text{F}} =\hat{\text{T}}_{1,2}- m_2\hat{g} = -(m_1 + m_2)\hat{a}$

Can anybody explain to me why they are using the sum of the masses as the net force acting on $m_2$?

 

[BvU]

Perhaps the a you calculate is the desired acceleration of mass 1 ? And how does the acceleration of mass 1 come about ?

I do have difficulty believing the book answer too, since $m_1\rightarrow 0$ begets g which I would say is a 100 % overestimate.

 

[matineesuxxx]

Perhaps the a you calculate is the desired acceleration of mass 1 ? And how does the acceleration of mass 1 come about ?

What exactly do you mean? Also, if there is a single mass, m, attached to a string with a downward acceleration of g/2, then then tension in the rope is mg/2; that, as well as your insight, leads me to believe their answer is wrong. However it may not be..

 

[TSny]

so I end up with $\hat{a} = \frac{\hat{g}}{2}$

I agree with your answer a = g/2.

 

[matineesuxxx]

I agree with your answer a = g/2.

Thanks for the confirmation. I am growing very displeased with this textbook due to the number of mistakes.. Take this for example; a question asking how fast you can spin a disk with a mass sitting on it before the mass slides off.

Part b) Find an expression for the tension in the string.

What string? Where is the string?? Not what i had in mind when I learned about "fictitious forces"

 

[TSny]

What string? Where is the string?? Not what i had in mind when I learned about "fictitious forces"

 

[BvU]

...spinning disk..
What string? Where is the string?? Not what i had in mind when I learned about "fictitious forces"

Nothing so good for building self esteem as improving on a textbook..., but:

I might agree with you on the spinning disk question being mistaken -- if that is the full tekst of the question. Is it ? Anyway, re the latter part of your post:

Fictitious forces are just that: ficitious. Most of the time. No need for them here. If we disregard snapping ropes (strings...), then m₁ and m₂ experience the same acceleration (it's a string, not a spring): $\ y_1- y_2\$ is a constant. Hence $\ \dot y_1 = \dot y_2\$ and $\ \ddot y_1 = \ddot y_2$.

Check it out: $\ T_{12} = {m_2g\over 2}\$ is a given. $\ \ddot y_1 = \ddot y_2 = g/2 \$ is what I claim. For $m_1$ the force balance is then $m_1 g/2 = m_1 g + m_2 g/2- T_1\$, or $T_1 = (m_1+m_2)\, g/2$. The two balls behave as a whole.

Not exciting at all; nothing fictitious needs to be brought in!