# Total energy loss due to resistance during circular motion

• furor celtica
In summary, we are given a problem involving a tube bent into a semicircle and a steel ball released into it. The ball experiences a constant resistance throughout its motion and comes to rest at points B and C. We are asked to show that the resistance is equal to (3mg)/(5π) and to find the speed of the ball at points B and C. We are also asked to show that 0.6α = π(cosα - 0.7) and to explain why the ball will continue to oscillate after reaching point C if α > arcsin(0.6/π). The solution involves using the total distance covered, not just the vertical distance, to calculate the work done by the resistance

## Homework Statement

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?

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furor celtica said:
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?
Use the total distance covered, not the vertical distance. (R opposes the motion of the ball.)

how do I calculate that?

furor celtica said:
how do I calculate that?
Just calculate the arc length in terms of the radius r. You have the angle.

Alright so that's figured, thanks for your help.
However, the problem doesn't stop there. Please inform me if I need to start a new thread for the remainder of the problem.

b. Find the speed of the ball at the lowest point of the tube
(i) as it moves from A to B
(ii) as it moves from B to C
c. Show that 0.6 (alpha) = pi((cos(alpha)) - 0.7)
d. Explain why the ball will continue to oscillate after reaching C if alpha > arcsin (0.6/pi)

I had no problem with all the other problems except for d.
By the way the answer to b. are (i). sqrt(7gr/5) and (ii). sqrt(3gr/5)

However I'm having a hard time with d.
Already the wording is troublesome, does 'continue to oscillate' mean 'continue to move' or 'cross over to other other quarter-circle again'?
In the first case I'm not sure what to formulate; I know the potential energy at C but what's the other part of the inequality? PE > 0? I don't think so.
In the latter case I tried solving it using PE at C > work done by R over distance alpha(r) but the answer is incorrect.

Suggestions?

sorry to bump but...bump
I've posted this on another physics help site as well but the helpers haven't been able to figure it. fancy a challenge (or something)?

furor celtica said:
However I'm having a hard time with d.
Already the wording is troublesome, does 'continue to oscillate' mean 'continue to move' or 'cross over to other other quarter-circle again'?
They want to know if it will continue to move. Hint: Consider the forces acting on the ball when it reaches C.

hmm. On another site the helpers said continuing to oscillate would mean traveling back distance alpha(r). But all my calculations (and theirs) based on this come up with nothing, so maybe you have a point.
But the problem with your method is finding the directions of the forces acting on the bead, i.e. gravity and R, and how would they tie in with angle alpha?

furor celtica said:
hmm. On another site the helpers said continuing to oscillate would mean traveling back distance alpha(r). But all my calculations (and theirs) based on this come up with nothing, so maybe you have a point.
They want to know if when it reaches point C does it start to swing back down (continue to oscillate) or just sit there.
But the problem with your method is finding the directions of the forces acting on the bead, i.e. gravity and R, and how would they tie in with angle alpha?
Make use of the results from the first part of the problem.

hmm so:
mg(sinα) > R
mg(sinα) > (3mg)/5π
α > 0.6/π

Got that.

How would I got about determining if this condition would be satisfied?

ok final bump, i promise
i just need help finishing this!
so i need to find out is alpha will satisfy the inequality, yes or no. I'm not asked to give alpha's value, i don't even know if that's possible. but how do i proceed?

You know the ball starts from point B, so figure out alpha based on that.

## 1. What is total energy loss due to resistance during circular motion?

The total energy loss due to resistance during circular motion is the amount of energy that is dissipated or converted into other forms, such as heat or sound, as an object moves in a circular path.

## 2. How is total energy loss calculated?

Total energy loss can be calculated by finding the difference between the initial energy of the system and the final energy of the system after the motion has occurred. This can be determined using the equation: ΔE = Ei - Ef, where ΔE is the change in energy, Ei is the initial energy, and Ef is the final energy.

## 3. What are some factors that affect total energy loss during circular motion?

The main factors that affect total energy loss during circular motion are the resistance force and the speed of the object. The greater the resistance force and the faster the object is moving, the higher the total energy loss will be.

## 4. How does resistance affect the total energy loss during circular motion?

Resistance, or the force that opposes motion, plays a significant role in the amount of energy lost during circular motion. The greater the resistance force, the more energy will be lost as the object moves in a circular path.

## 5. How can total energy loss due to resistance be reduced during circular motion?

Total energy loss due to resistance can be reduced by minimizing the amount of resistance force acting on the object. This can be achieved by using materials with lower friction coefficients, lubricating surfaces, or reducing the speed of the object.