Horizontal circular motion problem, what am doing wrong?

[SherlockLCooper]

Homework Statement

The County Fair Swing carries the mass of riders and chairs in a circular path in a horizontal plane while suspended by cables or chains. Let's assume that:

Each chair with riders is supported by a single cable

The tension in the cable equals 2 x the total weight riders and chair

The speed of the center of mass at the end of the cable is 10 m/s.

Determine the radius of the circular path in meters.Selected Answer:
50

Homework Equations

T=mv^2/r

The Attempt at a Solution

T=mv^2/r
so r=10^2/2
100/2=50
50=r

 

[Buzz Bloom]

Hi Sherlock:

I am not sure I understand the problem statement correctly, but it seems to me you are assuming that the chains are horizontal. I visualize a different picture with the chains at an angle. If that is correct, then the tension has a horizontal and a vertical component.

Hope this helps.

Regards,
Buzz

 

[jbriggs444]

If the cables are purely horizontal, what force supports the chairs and riders against gravity?

[i.e. -- what Buzz said]

 

[SherlockLCooper]

this is all the information I was given, I imagine the angle too, but I have no value for theta. is that something I can solve for with the information given?

 

[Buzz Bloom]

Hi Sherlock:

The vertical component of the tension balances what force? Visualize a right triangle with the chain as the hypotenuse.

The tension in the cable equals 2 x the total weight riders and chair

Regards,
Buzz

 

[BvU]

The tension in the cable equals 2 x the total weight riders and chair

What do you do with that information ?

T=mv^2/r so r=10^2/2

Do I read T/m = 2 here ? I always thought weight = mg so I am missing a factor g -- not to mention the dimension of g !

Oh, and -- in your relevant equation -- what do you mean with $\vec T$ ?

 

[Chestermiller]

Sherlock,

Have you drawn a free body diagram for the combination of rider and chair, or do you feel like you have advanced beyond the point where you need to use free body diagrams?

 

[BvU]

Well now, all this response and not even one -- let me make up for that: !

 

[SherlockLCooper]

t=tension. and I have drawn diagrams, and i just, tried v^2/2g, and it was also incorrect, am i using the right Equations?

 

[SherlockLCooper]

Well now, all this response and not even one -- let me make up for that: !

thanks!

 

[Nidum]

Have you drawn a free body diagram for the combination of rider and chair

It is always much easier to sort these types of problem out if there is something to look at .

Mark in the forces which you think are acting and any other information you consider relevant .

 

[Chestermiller]

t=tension. and I have drawn diagrams, and i just, tried v^2/2g, and it was also incorrect, am i using the right Equations?

Let's see what your free body diagram looks like

 

[SherlockLCooper]

does this help

 

[Buzz Bloom]

Hi Sherlock:

Your diagram is a good start, but you need to add the horizontal and vertical components of the tension.

Regards,
Buzz

 

[Chestermiller]

does this help

Yes. OK If T is the tension, what are its components in the x and y directions, in terms of the angle the rope makes with the horizontal?

 

[SherlockLCooper]

I want to say
(T Cos(theta)i, T sin(theta)j) where the y component would be T Sin(theta)=mg..?
so t/mg= 2, so sine (0) =0, so it is in the horizontal plane, the alternative T Cos(0)= T*1= T.

 

[SherlockLCooper]

so all the velocity would be in the horizontal plane?

 

[Buzz Bloom]

Hi Sherlock:

If you visualize the right triangle, you are given the facts regarding the vertical component, and the tension is along the hypotenuse. You don't need the angle to calculate the horizontal component. You can use the Pythagorean theorem.

Regards,
Buzz

 

[Chestermiller]

I want to say
(T Cos(theta)i, T sin(theta)j) where the y component would be T Sin(theta)=mg..?
so t/mg= 2, so sine (0) =0, so it is in the horizontal plane, the alternative T Cos(0)= T*1= T.

If T = 2mg, then $2mg \sin\theta=mg$, then what is $\sin \theta$ and what is $\theta$?

 

[BvU]

Pity T is missing in the side view in post #13...

 

[SherlockLCooper]

30 degrees!

 

[Chestermiller]

30 degrees!

Good. Now let's see your force balance in the horizontal direction. There is only one force acting in the horizontal direction.

 

[SherlockLCooper]

(2mg cos 30)/m=a=v^2/r
m cancels so v^2/2gcos30=r

 

[SherlockLCooper]

(100)/(2)(9.8)((3^(1/2))/2)=5.89 meters

 

[SherlockLCooper]

ill try this and let you know! Thank you so much!

 

[SherlockLCooper]

it worked!

 

[Chestermiller]

it worked!

The key learning from all of this is reconfirmation of the importance of always using free body diagrams.