The problem is as follows:
The figure from below shows two brass cylinders sliding through a wire as a guiding line to hit the ground. The cylinders are released from point $P$. Assuming that the friction between the wire and the cylinders is negligible. Find the relation between the elapsed time required for each cylinder to get the ground. $\frac{t_b}{t_a}$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\tan^2\theta\\ 2.&\csc\theta\\ 3.&\sec\theta\\ 4.&\cos^2\theta\\ 5.&\sin\theta\\ \end{array}$
My attempt to solve this problem is sketched in the diagram from below:
I'm assuming that the length from the ground to $P$ for the cylinder labeled $A$ is $r$. Then this would account for the height as $r\sin\theta$
for cylinder $B$:
$y=y_o+v_{oy}t-\frac{1}{2}gt^2$
The intial conditions here from point $P$:
$y_o=r\sin\theta and v_{oy}=0$
becuase it mentions that it is released. It does not specifically mention but I think the intended meaning was that is from rest.
so to get the time $y=0$
$0=r\sin\theta-\frac{1}{2}gt^2$
$t_b=\sqrt{\frac{2r\sin\theta}{g}}$
Now it comes the tricky part which I'm confused. Does it mean for the cylinder on $A$ does it has horizontal acceleration?. I think it has because it might be influenced from the acceleration due gravity.
Since the only acceleration is acting vertically the horizontal acceleration would be:
$a_x=g\sin\theta$
But is this part correct?. Is it okay to assume that's the horizontal acceleration?. If so why does it appear?. Is it because of the wire?.
Then:
The speed at the end when the cylinder gets the ground can be found from using the Energy theorem:
$E_p=E_k$
$mg(r\sin\theta)=\frac{1}{2}mv^2$
Hence:
$v_b=\sqrt{2rg\sin\theta}$
But the energy theorem accounts for total velocity thus the vertical component would be:
$v_y=-\sqrt{2rg\sin\theta}\sin\theta$
Thus to get the time can be found from using the vertical motion equation:
$v_{f}=v_o-gt$
$t=\frac{v_{o}-v_{f}}{g}$
Then inserting the speeds:
$t_a=\frac{0-(-\sqrt{2rg\sin\theta}\sin\theta)}{g}$
$t_a=\frac{\sqrt{2rg\sin\theta}\sin\theta}{g}$
$t_a=\sqrt{\frac{2r\sin\theta}{g}}\sin\theta$
Then by dividing $\frac{t_b}{t_a}$ results into:
$\frac{t_b}{t_a}=\frac{\sqrt{\frac{2r\sin\theta}{g}}}{\sqrt{\frac{2r\sin\theta}{g}}\sin\theta}$
$\frac{t_b}{t_a}=\frac{1}{\sin\theta}=\csc\theta$
Which I believe the answer should be $\csc\theta$ but my book say that the correct answer is the fifth option or $\sin\theta$. What could be going on here?. Can someone help me here?.
Is it that am I understanding this not correctly?. Is my approach correct?. Can someone help me about the horizontal component for the cylinder?. Does it has an acceleration?. Is this correct?.
Think about the forces acting on cylinder $A$. You have gravity (going down) and a force from the wire, acting perpendicular to the wire (like the normal force from an incline plane). You can decompose these forces in two directions, one along the wire and one perpendicular to it. Since the cylinder is not moving in a direction perpendicular to the wire, it means that the normal force and the component of gravity along the perpendicular direction cancel out. In the direction along the wire you only have a component of gravity $mg\sin\theta$, so the acceleration is $g\sin\theta$. Then the time to travel a distance $r$ is $$t_a=\sqrt{\frac{2r}{g\sin\theta}}$$so $$\frac{t_b}{t_a}=\frac{\sqrt{\frac{2r\sin\theta}{g}}}{\sqrt{\frac{2r}{g\sin\theta}}}=\sin\theta$$