\begin{problem*}{5.52}
An amusement park ride consists of a rotating circular platform
$d=8.00\U{m}$ in diameter from which $m=10\U{kg}$ seats are suspended
at the end of $l=2.50\U{m}$ massless chains (Fig. P5.52).  When the
system rotates, the chains make an angle of $\theta=28.0\dg$ with the
vertical.
\Part{a} What is the speed of each seat?
\Part{b} Draw a free-body diagram of a $m_c=40.0\U{kg}$ child riding
in a seat, and find the tension in the chain.
\end{problem*} % problem 5.52

\begin{solution}
\Part{a}
We will eventually use $v=\sqrt{a_c r}$ as we have in all the other
problems in this homework assignment to find $v$.

First, we need to find the radius $r$ of the path that the seat takes
around the ride.
\begin{equation}
 r=\frac{d}{2} + l \sin \theta
  =(4.00 + 2.50\sin 28.0\dg)\U{m}
  =5.1736\ldots\U{m}
\end{equation}

Now we need to find the centerward acceleration $a_c$.  Drawing a free
body diagram of our seat, we see that the only forces acting upon it
are the tension \vect{T} and gravity $\vect{F}_g$.  We know that the
seat does not rise or fall in the vertical (\vect{y}) direction, so
summing the forces we have
\begin{align}
 \sum F_y &= T \cos \theta - mg=m a_y=0 \\
  T &= \frac{mg}{\cos\theta} \label{eqn.T}\\
 \sum F_c &= T \sin \theta 
          =mg\tan\theta
          =m a_c \\
 a_c &= g\tan\theta
    =9.8\U{m/s}^2 \cdot \tan 28.0\dg
    =5.2108\ldots\U{m/s}^2 \label{eqn.ac}
\end{align}
So
\begin{equation}
 v=\sqrt{a_c r}
  =\sqrt{g \tan \theta \cdot (\frac{d}{2} + l \sin \theta)}
  =\sqrt{ 5.2108\ldots\U{m/s}^2 \cdot 5.1736\ldots\U{m}}
  =5.19\U{m/s}
\end{equation}

\Part{b}
Our free body diagram with a child in the seat will be the same as our
diagram from \Part{a} but with a new mass $m'=m + m_c=50\U{kg}$.

Before we find the tension in the chain, we should check to see if the
chain angle changes.  The angular velocity $\omega=v/r$ does not
change when people get into the seats (because they are of negligible
mass compared to the platform), so we can relate our new velocities
$v'$ and $r'$ using the same $\omega$ that we had in \Part{a}.
\begin{equation}
 \omega=\frac{v'}{r'}
       =\frac{v}{r}
       =\frac{\sqrt{ g r \tan \theta}}{r}
	= \sqrt{\frac{g\tan\theta}{r}}
       =1.00\U{rad/s}
\end{equation}
Not that the numerical value is important, just that it is a constant.
We can plug $v'=\omega r'$ into our centerward acceleration equation
\begin{equation}
 a_c'=\frac{v'^2}{r'}
    =\frac{\omega^2 r'^2}{r'}
    =r' \omega^2
\end{equation}
And applying this to eqn. \ref{eqn.ac}(which hasn't changed except for
the need to substitute primed variables)
\begin{align}
  a_c' &= g \tan\theta'=r' \omega^2 \\
  g \tan\theta' &= \left( \frac{d}{2} + l \sin\theta' \right) \omega^2
\end{align}
The only unknown in this equation is $\theta'$, but the equation is
analytically unsolvable.  We know $\theta'=28.0\dg$ is one solution,
because there are no masses in this equation, so is must also hold for
case \Part{a}.  Then we have to decide if there will be any other
solutions.  We know intuitively that any solutions will have $0\dg <
\theta' < 90\dg$.  Considering the $\sin$ and $\tan$ functions on that
interval, we see that $\sin\theta'$ is concave down and continuous
over the entire interval, and that $\tan\theta;$ is concave up and
continuous over the entire interval.  Therefore, the left hand side of
this equation only equals the right hand side at a single value of
$\theta'$ so our $28.0\dg$ solution is unique.  If this doesn't make
sense to you, you can graph the right and left hand sides to check.

Having proved that $\theta'=\theta$ we can move on to solve for the
tension.  Using eqn \ref{eqn.T}.
\begin{equation}
 T'=\frac{m' g}{\cos \theta'}
   =\frac{50\U{kg} \cdot 9.8\U{m/s}^2}{\cos 28.0\dg}
   =\ans{555\U{N}}
\end{equation}

As far as grading is concerned I will accept anything where you did
any of the following:
\begin{itemize}
\item assumed $\theta$ didn't change (skipping the whole $\theta'=\theta$ step)
\item assumed $\omega$ didn't change, and you went on to show
  $\theta'=\theta$ is a valid solution (skipping the uniqueness step).
\item assumed $\omega$ didn't change, and proved that $\theta'=\theta$
  is valid and unique.
\end{itemize}
\end{solution}
