A tube is bent into the form of a semicircle with centre O and radius r. It is fixed in a vertical plane with its diameter horizontal, as shown in the diagram. A steel ball is held at one end A, and released into the tube. Throughout its motion the ball experiences a resistance of constant magnitude R. The ball first comes to rest at B, where OB makes an angle of π/3 with the vertical. It then runs back down the tube and next comes to rest at C, where OC makes an angle α with the vertical.
Show that R = (3mg)/(5π)
The Attempt at a Solution
Alright so the problem I immediately encountered in trying to solve this is that I am unsure of how to find formulate the work done by R. Is this expressed over the total circular distance covered, or simply vertical distance?
I tried using the latter method and came up with a wildly incorrect answer, but my work looked like this:
(Potential energy at A) – (potential energy at B) = (work done by R over the vertical distance r+(r-x))
I ended up getting R = (mg(sin(π/6)))/(12 – sinπ), which obviously is not even close to correct.
I assuming it is incorrect to assume R is working vertically. But how else do I formulate the energy loss caused by R? Do I even need this expression to solve the problem, or is there an alternative method?
1.8 KB Views: 366