# Dynamics of Circular Revolution

1. Feb 2, 2008

### Arejang

[SOLVED] Dynamics of Circular Revolution

1. The problem statement, all variables and given/known data
In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure View Figure , one of which is horizontal. The seat swings in a horizontal circle at a rate of 33.1 rev/min

If the seat weighs 252 N and a 831 N person is sitting in it, find the tension in the horizontal cable.

If the seat weighs 252 N and a 831 N person is sitting in it, find the tension in the inclined cable.

2. Relevant equations

If I read the problem right, this should be the formula used

(4pi^2*R)/T^2

3. The attempt at a solution

I first found the a_rad by plugging in 7.5m for R and 60sec/33.1rev for T. Afterwards, I convert the given Newtons of the chair and the person from problem to kilograms by dividing by 9.8 m/s^2. Once I calculated the mass of both person and chair, I added them together and multiplied by my acceleration from earlier to get the amount of Newtons for the tensions for the cables. A big question is, I'm pretty sure trig absolutely must play a role in it due to the angle of 40 degrees given, but I'm not sure whether it should be 7.5*sin(theta) or 7.5*cos(theta).

(4pi^2*7.5*sin(40))/(60/33.1)^2

Before I did anything with this resultant acceleration, I first added two weights of the seat and the person. Then I divided the sum by 9.8 m/s^2 to get the mass in kilograms. Then I multiplied the resultant kilograms to the acceleration I calculated from above to get 64000 N. This was for the horizontal tension on the cable. Please let me know if I had done anything wrong. Thanks!

#### Attached Files:

• ###### YF-05-58.jpg
File size:
9.3 KB
Views:
632
Last edited: Feb 2, 2008
2. Feb 2, 2008

### Arejang

Bump! Did I use the correct formula and everything? Please let me know if I'm approaching the problem from the right angle. ^_^

3. Feb 2, 2008

### Littlepig

i guess an explanation to the problem would be good as

4. Feb 2, 2008

### Arejang

Sorry, picture is now uploaded. :)

5. Feb 2, 2008

### Arejang

Bump, Please let me know if I can clarify the problem in any way, but I'm pretty sure you have the entire problem now. If any part of it is unclear, though, let me know.

6. Feb 3, 2008

7. Feb 3, 2008

### Staff: Mentor

Rather than use some canned formula (what's that formula for, anyway?), start from first principles. Identify the forces acting on the "person+seat" and apply Newton's 2nd law. Analyze the vertical and horizontal components separately and then combine the two equations.

8. Feb 3, 2008

### Arejang

So the forces acting on the person+seat are the horizontal cable and the inclined cable, their weight (252 N and 831 N), and gravity. A period of 33.1 rev/min is given, once converted to seconds (60 sec/ 33.1 rev), you have for every one sec, 1.81 revolutions, or 1.81 rev/sec.

Using Newton's Second Law, the net force is equal to the product of the mass and acceleration of a given object. The net force is also divided into horizontal and vertical components. So the product of the mass and acceleration on the horizontal component and vertical component are treated like two separate coordinates. In this problem, there does not seem to be any vertical acceleration.

So to solve the net force on a given body, we need to know both the mass of the body in question, (which in this case is the weight of the person + seat), and its acceleration as it revolves.

The horizontal acceleration is toward the center of the circle. Since the acceleration I need to figure out is circular, in other words, a radial acceleration, the equation I used above is used to calculate the acceleration of a body moving in a uniform circular motion. I forgot to label everything, which would probably explain a lot more.

(4pi^2*7.5meters*sin(40 degrees))/(60seconds/33.1revolutions)^2

The result should give an acceleration of 57.9 m/s^2.

Applying Newton's 2nd law, of ΣFnet= m*a, I converted the Newtons to an easier to use mass, kg by dividing each weight by 9.8 m/s^2. Then I added the two weights together, assuming that the tension of the horizontal cable would have to support both weights simultaneously. Then I multiplied the sum weights with the acceleration I derived from above.

I'm pretty sure I'm misunderstanding the concept of Newton's Law somehow, but I just can't figure out where. The work I just did was only for the horizontal component of the two bodies, but I think that even that wasn't done with a solid understanding of the 2nd law. Please tell me if you see any flaw in my calculations.

Last edited: Feb 3, 2008
9. Feb 3, 2008

### Staff: Mentor

Their weight is gravity.
Careful when converting to rev/sec.

OK.

OK.

You are using $a_c = \omega^2 r = 4 \pi^2 f^2 r$. Why did you multiply by sin(40)?

Draw yourself a diagram showing all the forces on the combined mass including their directions. Call the tensions T1 (inclined) and T2 (horizontal). What's the net force in the horizontal direction? Set that equal to ma.

What's the net force in the vertical direction? What must that equal?

You'll end up with two equations, which will allow you to solve for the two unknown tensions.

10. Feb 3, 2008

### Arejang

K, My mistake on the revolutions per second. It seems I have converted revolutions per 60 seconds. There should be 0.552 revolutions per second.

If not here, I don't see where the 40 degrees would have any significance. Again, I'm saying this not to insist I'm right, but that I'm sincerely lost as to where to go with this.

I've been doing it the other way around. The mass and acceleration seemed like they would be the first targets to solve, then the net force could be identified.The only other forces given are the weights of the two bodies. I'm not sure I follow you.

The net force in the vertical direction should be equal to the weight of the two masses as there is no vertical acceleration. But since the bodies aren't moving in any vertical direction, shouldn't the total net force equal zero?

Would it be these? $$\Sigma$$$$F^{}_y{}$$= n + (-mg)=0

and $$\Sigma$$$$F^{}_x{}$$=$$ma_x{}$$?

These would be the most basic formulas, but I'm still unsure how to apply these in circular motion.

I was also wondering if I converted the weight in Newtons into kilograms correctly. I'm going to use the formula you gave me if I did. I understand w to mean weight, which would be the newtons of the masses converted into kilograms (which, from 252 Newtons of the seat and 831 Newtons of the person would be 25.7 kilograms and 84.8 kilograms respectively), the r to be the radius, which in this case is 7.5 meters, and the force, f, to be the weight of the two bodies in newtons.

Edit: Sorry for my horrible usage of this latex program. I hope that it's still somewhat decipherable.

Last edited: Feb 3, 2008
11. Feb 4, 2008

### Staff: Mentor

OK.
What matters in calculating the centripetal acceleration is the radius of the circular path. That radius is given as 7.5 m. No need for any trig.

That angle will come in handy later when you are analyzing the components of the tension from the inclined cable.
Treating the masses as one combined mass, there are 3 forces acting: Weight (acting down), tension 1 (acting along the incline), and tension 2 (acting horizontal). Find the horizontal and vertical components of these three forces.
The net force in the vertical direction is the sum of the vertical components of all 3 forces. Add them up! Since there's no vertical acceleration, that sum must equal 0. That's one equation that you need.

To get the second equation, add up the horizontal components of all 3 forces. Since there is a horizontal acceleration, that sum must equal ma.
Redo this. There's no normal force (assuming that's what "n" stands for), but there is a vertical component of one of the tensions.
Nothing wrong with that. Add up those horizontal components.
You'll use the formula for centripetal/radial motion to find the acceleration.
Your conversion to kg is just fine. Weight = mg, so m = w/g.

12. Feb 4, 2008

### Arejang

K, I went back to school, and got some feedback from my friends. Tell me if I did this right.

If we label the inclined cable as $$T_{2}$$, we solve for it first. This is done by adding the two weights as newtons together, then dividing by the cos(40).
So,

(252N+831N)/cos(40)=1413.7 or 1414

So the centripetal acceleration formula is as follows: $$a_{rad}=$$$$\frac{v^{2}}{R}$$

If $$a_{rad}$$ was expressed in terms of period T it can be written as so:

$$T=\frac{2\pi R}{v}$$

In terms of the period, $$a_{rad}$$ is

$$a_{rad}=$$$$\frac{4\pi^{2}R}{T^{2}}$$

R for the radius of 7.5 meters, and T being 1.81 seconds per revolution. Plugging and chugging the numbers gives 90.38 $$m/s^{2}$$.

Multiplying this number with 110.5 kg (the weight of the two bodies, now converted into kg) should give you the force 9985.885N

And apparently $$T_{1}$$ must support the weight of both itself and $$T_{2}$$. So you have, by calculating the sum of both forces, and setting it equal to the N we solved above, 9985.885N, the tension for $$T_{1}$$. So laid out this should be what it looks like

9985.885N=$$T_{1}$$+$$T_{2}$$*sin(40)

9985.885N=$$T_{1}$$+1414cos(40)

Solving for T, you get

9985.885-1414cos(40)=8902.698

I'm not sure if all the calculations or the theory is correct. The online homework program stated I was a little more than 100 off, but accepted my answer anyway. I'm not sure why, but curve me in the right direction with this problem, so I can ensure myself that I know what I'm doing is right, and why.

13. Feb 4, 2008

### Staff: Mentor

Looks good. This is just applying Newton's 2nd law to the vertical direction:
$$\Sigma F_y = 0$$

$$T_2 \cos(40) - W = 0$$

$$T_2 = W/\cos(40)$$

Looks good.

OK. That's the net force in the horizontal direction.

The way to look at it is using Newton's 2nd law again:

$$\Sigma F_x = ma_c$$

$$T_2 \sin(40) + T_1 = ma_c$$

$$T_1 = ma_c - T_2 \sin(40)$$

Good.

Why did you change sin to cos?

Correct this.

I only see one mistake.

14. Feb 4, 2008

### Arejang

Awesome. The answer I got from correcting the cos to sin, I resulted with 9077. The answer desired was 9050. Though it wasn't the exact number, it was a lot closer to the right answer than 8902 (which results in a 148 difference).

A couple last questions I have to ask before I mark this solved.

What are fundamental questions that must be asked before attempting to tackle a Newton's 2nd Law problem? My biggest difficulty with this problem especially was figuring out where to begin. But I found this same roadblock on other similar problems as well. Though I understand the law in formulaic form is F=ma the problem doesn't always come out clearly enough for me to identify, "Oh, this is a 2nd law problem," or even "Because this is a 2nd law problem, I must need this kind of structure so using the info I have, I must work to result in the desired ideal structure, so that the formula would work".

In either case, I'm lost as to how I must tackle the problem. So I guess what I want to say is this: what are some key words that makes a 2nd law problem stick out like a sore thumb? And what are the immediate things you identify first before tackling it?

15. Feb 5, 2008

### Littlepig

I guess generally is about what is asked. You read the problem the first time, with the question, and then, reread the problem having in mind what is your objective.
So, when you are rereading, you start to "see what puzzle pieces you need"(putting equations you will probably need, data given etc), then, you start putting all your pieces together(joining equations you know with the data given, manipulating) in order to fit piece by piece, till the puzzle is done.

This is to close problems(with a unique solution, or solution explicit)

16. Feb 5, 2008

### Staff: Mentor

clues to Newton

Whenever you given or asked to find forces or acceleration, that should give you a clue that Newton's 2nd law might be involved. In this problem they ask you to find the tensions--forces--that's one tip off. Another clue is that circular motion is involved, which should tell you that something is accelerating. (And whenever equilibrium is mentioned or can be deduced, that will tell you that the acceleration (and net force) is zero.)

17. Feb 5, 2008

### Arejang

Thanks a lot. Hopefully, I'll be better equipped to find solutions to these kinds of problems. ^_^;