# Object is dragged across a semicircle's perimeter

• Const@ntine
In summary: Cartesian coordinates. It is easiest to answer both questions in terms of theta. That integral solution you found is exactly what you need. Now you just need the initial and final theta.
Const@ntine

## Homework Statement

A small object with mass m is dragged without friction to the highest point of a semicircle with a radius of R, by a weightless rope.

a) If the magnitude of his velocity is constant, then his acceleration that is parallel to the semicircle, is zero. Prove that F = mgcosθ

b) Integrate the following swiftly W = ∫ F*dr, to calculate the work that is produced when the object is moved from the highest point, to the lowest.

## The Attempt at a Solution

(a) So, first up I set the direction of the force F as the Y axis, and then created the net force (I don't know the proper term yet, but you get the gist of it from the pic). Like this:

Thus: ΣFy = mat = 0 <=> F = Fgy = Fgcosθ = mgcosθ

(b) Now's the part where I can't quite figure it out. I didn't do much intergals at school, so it's pretty new to me, and I haven't reached that part in my math book. I glanced at the back and found the basic ones (eg ∫cosθdθ = sinθ + c), but I've never come across an intergal with both x & y coordinates. I took a look at other examples from the solution manual, but only one other problem was intergals, and it's quite different.

The book's answer is "mgR", and I kinda sort of found the same, but from lowest to highest, not the opposite, like the book's asking. If anyone could help me with the intergal, I'd appeciate it!

Darthkostis said:
b) Integrate the following swiftly W = ∫ F*dr, to calculate the work that is produced when the object is moved from the highest point, to the lowest.

The book's answer is "mgR", and I kinda sort of found the same, but from lowest to highest, not the opposite, like the book's asking. If anyone could help me with the intergal, I'd appeciate it!
I'm not sure what the problem setter had in mind. From the instruction to integrate F, I would say highest to lowest is a typo. Must have meant lowest to highest. There is no information about F in a highest to lowest phase.
The "swiftly" is suspicious. Might mean that you are not supposed to do the integral at all, just use work conservation and write down mgr straight away. Hard to tell.

haruspex said:
I'm not sure what the problem setter had in mind. From the instruction to integrate F, I would say highest to lowest is a typo. Must have meant lowest to highest. There is no information about F in a highest to lowest phase.
The "swiftly" is suspicious. Might mean that you are not supposed to do the integral at all, just use work conservation and write down mgr straight away. Hard to tell.

Yeah, I just took it as a typo as well. From lowest to highest I get the book's result, so that's probably it.

Darthkostis said:

## Homework Statement

A small object with mass m is dragged without friction to the highest point of a semicircle with a radius of R, by a weightless rope.

a) If the magnitude of his velocity is constant, then his acceleration that is parallel to the semicircle, is zero. Prove that F = mgcosθ

b) Integrate the following swiftly W = ∫ F*dr, to calculate the work that is produced when the object is moved from the highest point, to the lowest.

## The Attempt at a Solution

(a) So, first up I set the direction of the force F as the Y axis, and then created the net force (I don't know the proper term yet, but you get the gist of it from the pic). Like this:

Thus: ΣFy = mat = 0 <=> F = Fgy = Fgcosθ = mgcosθ

(b) Now's the part where I can't quite figure it out. I didn't do much intergals at school, so it's pretty new to me, and I haven't reached that part in my math book. I glanced at the back and found the basic ones (eg ∫cosθdθ = sinθ + c), but I've never come across an intergal with both x & y coordinates. I took a look at other examples from the solution manual, but only one other problem was intergals, and it's quite different.

The book's answer is "mgR", and I kinda sort of found the same, but from lowest to highest, not the opposite, like the book's asking. If anyone could help me with the intergal, I'd appeciate it!

You shouldn't convert to Cartesian coordinates. It is easiest to answer both questions in terms of theta. That integral solution you found is exactly what you need. Now you just need the initial and final theta.

Cutter Ketch said:
You shouldn't convert to Cartesian coordinates. It is easiest to answer both questions in terms of theta. That integral solution you found is exactly what you need. Now you just need the initial and final theta.

I'm not sure how though. As I've said, I'm not that familiar with intergals, so I'm not sure how to get it to work. I figure that at the highest point, theta is 90 degree, and at the lowest, it's 0 degree. But I'm not exactly sure how to handle the intergal.

Cutter Ketch said:
Well, you have the antiderivative, so now you just need to know about definite integrals.

I know how the basic version works (only one kind of variable, say, x), and I have some experience with more complex types, but I don't know how to connect the angle (theta) and the displacement (r). The best I've got is:

W = ∫Fdr = ∫HighestLowestmgcosθdy = ∫900mgcosθRdθ = mgR∫900(sinθ)'dθ = mgR[sinθ]900 = mgR(sin90 -sin0) = mgR

I'm not sure if that's correct though.

Cutter Ketch
Darthkostis said:
I know how the basic version works (only one kind of variable, say, x), and I have some experience with more complex types, but I don't know how to connect the angle (theta) and the displacement (r). The best I've got is:

W = ∫Fdr = ∫HighestLowestmgcosθdy = ∫900mgcosθRdθ = mgR∫900(sinθ)'dθ = mgR[sinθ]900 = mgR(sin90 -sin0) = mgR

I'm not sure if that's correct though.
That all looks fine, but it is a little confusing to use r and y as variable names with such non-standard meanings. You could use ds as a generic displacement along a path, and write ds=Rdθ.

Cutter Ketch
haruspex said:
That all looks fine, but it is a little confusing to use r and y as variable names with such non-standard meanings. You could use ds as a generic displacement along a path, and write ds=Rdθ.
I had the y-axis in my mind, hence why I used that symbol. But yeah, your way is better.

Darthkostis said:
I know how the basic version works (only one kind of variable, say, x), and I have some experience with more complex types, but I don't know how to connect the angle (theta) and the displacement (r). The best I've got is:

W = ∫Fdr = ∫HighestLowestmgcosθdy = ∫900mgcosθRdθ = mgR∫900(sinθ)'dθ = mgR[sinθ]900 = mgR(sin90 -sin0) = mgR

I'm not sure if that's correct though.

Yes, that's it. And better with Haruspex's suggested labeling.

Well, that settles it then.

Thanks for the help everybody!

## 1. What is the definition of "Object is dragged across a semicircle's perimeter?"

The term "Object is dragged across a semicircle's perimeter" refers to the action of moving an object along the curved boundary of a semicircle, either by pulling or pushing it.

## 2. What are some examples of objects that can be dragged across a semicircle's perimeter?

Examples of objects that can be dragged across a semicircle's perimeter include a pen, a toy car, a ruler, or any other small object that can be easily moved along a curved path.

## 3. What are the factors that affect the speed at which an object is dragged across a semicircle's perimeter?

The speed at which an object is dragged across a semicircle's perimeter can be affected by various factors such as the force applied to the object, the mass of the object, the surface of the semicircle, and the angle at which the object is dragged.

## 4. How does the direction of the drag affect the movement of the object across a semicircle's perimeter?

The direction of the drag can significantly impact the movement of the object across a semicircle's perimeter. If the object is dragged in a straight line, it will follow a curved path along the semicircle's perimeter. However, if the object is dragged at an angle, it will move in a more complex path, potentially resulting in a spiral-like movement.

## 5. What are some practical applications of dragging an object across a semicircle's perimeter?

Dragging an object across a semicircle's perimeter can have various practical applications, such as measuring the circumference of a circle, demonstrating circular motion in physics, or creating curved designs in art or architecture. It can also be used in sports, such as discus or shot put throwing, where the object is dragged in a circular motion before being released.

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