The problem is as follows:
A brass sphere is hanging from a ceiling as it is seen in the figure
from below. Assuming that at the position $A$ the speed is
$3\frac{m}{s}$ the bob swings and hits a bolt located in point $B$. As
a result the bob keeps swinging in the trajectory indicated by the
dashed line.
Using this information find the speed of the bob when it passes
through point $C$. You may assume that the acceleration due gravity is
$g=10\frac{m}{s^2}$.
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{3.6 meters per second}\\
2.&\textrm{2.8 meters per second}\\
3.&\textrm{4.2 meters per second}\\
4.&\textrm{1.8 meters per second}\\
\end{array}$
So far what I've attempted to do is to use the conservation of mechanical energy on this system as follows:
Assuming that the moment when the bob is at $A$ it will have the same energy as when it passes below the point $B$. I'm assuming that the reference for the potential energy is just below point $B$. I mean where the bolt is put in the wall.
$E_{kA}+E_{uA}=E_{kB}$
Well after doing this I'm getting:
$\frac{1}{2}mv^2_{A}+mgR=\frac{1}{2}mv^2_{B}$
Cancelling the mass and replaing the known quantities I'm getting:
$v^2_{B}=9+8$
$v^{2}_{B}=17$
Now in this second segment I'm assuming that the energy on point $B$ will remain the same as when it passes through point $C$.
Therefore:
$\frac{1}{2}mv^2_{B}=\frac{1}{2}mv^2_{C}+mgR$
But since R=0.4 this will make that the speed on $C$ to be also 3 meters per second.
What could it be wrong in this analysis?. Did I overlooked something?. Can someone help me to spot what did I do wrong?.
The more I look into this question, I think that this may have to do to work with the tension of the wire.
I have not used the fact that of the angle which was given. Should this be used?.
If I were to use the angle initially I could calculate the initial tension of the wire. I mean on point $A$.
But this tension will not be the same when the bob passes through point $C$. Perhaps does it exist a way to find it?.
Can someone guide me here?. I feel lost. Please try to include a step by step explanation so I can understand.
Best Answer
The diagram is deceptive: the bob is $ \ 1.2 \cos 60º \ = 0.6 $ m. below the pivot, which is higher than point C at $ \ 0.8 $ m. below the pivot. [The bob goes from 0.6 m. to 0.4 m. above the lowest point of the pendulum swing.] So the change in potential energy of the bob is $ \ -0.2 mg \ $ from point A to point C .
The speed at point C is then given by
$ \ \frac{1}{2} \ v^{2} \ = \ \frac{1}{2} · 3^{2} \ + \ (0.6 - 0.4) · g \ $ ;
this does give one of the available choices for the result.
Concerning the tension in the wire, while it is true that $ \ T \ $ throughout the swing varies, this force is always perpendicular to the circular arc of the swing. So in a simple pendulum, the tension does no work on the bob, and so does not enter into the kinetic energy - potential energy calculation. The tension force only serves to control the trajectory of the bob.