Help with uniform motion problem.

[thestar88]

Homework Statement

A 0.60 kg sphere rotates around a vertical shaft supported by two strings, as shown. If the tension in the upper string is 18 N, calculate the tension in the lower string.

Homework Equations

ƩF_R = ma_R

F_T = m * (v² / r)

I am not sure about this, but I also think that the equation to find the centripetal acceleration may come in handy as well.

The Attempt at a Solution

I am not quite sure how to approach this problem. I tried to use the same methods that were used for a similar problem found here: http://www.educator.com/learn/physics/physics-b/jishi/circular-motion-part-1.php?ex=1 (It is extra example #1), but I do not feel confident in this approach. Physics is not my strongest subject, so please forgive my utter cluelessness. My best guess for this problem is to perhaps solve for the acceleration of the top small triangle where String #1 is the hypotenuse, and then use that answer to solve for the tension of the bottom string.

Please correct me if I am wrong, and if you have any tools to help with understanding Uniform circular motion (or physics in general), please let me know. Thanks for the help!

 

[PhanthomJay]

Instead of looking in the horizontal direction , try looking in the vertical direction where there is no acceleration. Draw a free body diagram of the sphere and use Newton's 1st law.

 

[thestar88]

More help please...

Here is what I have done so far:
1. I found out the length missing side of the top triangle (the missing side is T1 or the hypotenuse of the triangle) using Pythagoras.
2. Given that the top triangle is a right angle triangle, I proceeded solve for the angles using the sin/cos properties of a right angle triangle.
2
that is all that I have. I have tried to Could you please clarify what you mean by the vertical direction? I know that this system will revolve horizontally (kind of like a tether-ball), so there will be 0 vertical velocity, which also means 0 vertical acceleration. Also, how can I use the 18 N given in the equation to find the tension of the lower string? How can I use the equation T₁sinθ + T₂cosθ = m × (v²/r) to find the T₂ when I don't know the velocity of the system?

 

[PeterO]

Here is what I have done so far:
1. I found out the length missing side of the top triangle (the missing side is T1 or the hypotenuse of the triangle) using Pythagoras.
2. Given that the top triangle is a right angle triangle, I proceeded solve for the angles using the sin/cos properties of a right angle triangle.
2
that is all that I have. I have tried to Could you please clarify what you mean by the vertical direction? I know that this system will revolve horizontally (kind of like a tether-ball), so there will be 0 vertical velocity, which also means 0 vertical acceleration. Also, how can I use the 18 N given in the equation to find the tension of the lower string? How can I use the equation T₁sinθ + T₂cosθ = m × (v²/r) to find the T₂ when I don't know the velocity of the system?

You can resolve the tension in the top string to produce a vertical and horizontal component.

The vertical component is part of the force supporting the bob (vertically). The horizontal component is part of the centripetal force acting on the bob.

The vertical and horizontal components of the tension in the bottom string supply "the rest" of those forces.

Theoretically you could take your pick of the horizontal or vertical requirements to solve this problem - but that would assume you know the size of the required centripetal force. You might be limited to considering the vertical - a hint given earlier.

 

[PhanthomJay]

Here is what I have done so far:
1. I found out the length missing side of the top triangle (the missing side is T1 or the hypotenuse of the triangle) using Pythagoras.
2. Given that the top triangle is a right angle triangle, I proceeded solve for the angles using the sin/cos properties of a right angle triangle.
2
that is all that I have. I have tried to Could you please clarify what you mean by the vertical direction? I know that this system will revolve horizontally (kind of like a tether-ball), so there will be 0 vertical velocity, which also means 0 vertical acceleration. Also, how can I use the 18 N given in the equation to find the tension of the lower string? How can I use the equation T₁sinθ + T₂cosθ = m × (v²/r) to find the T₂ when I don't know the velocity of the system?

Since you don't know the velocity, you cannot initially proceed in the horizontal direction. In the vertical direction, however, which is the y direction, you know that since there is no acceleration on that direction, the sum of all components of the forces on the sphere I am thst direction must add to 0. So what are the y components of the tension forces and weight, using trig to get the vertical leg component values?

 

[thestar88]

I have another question: Can I use the vertical acceleration (which equals 0) to solve for the acceleration of this system and then solve for the frequency (or rev/min) of the system?
Thanks again for the help.

 

[PeterO]

I have another question: Can I use the vertical acceleration (which equals 0) to solve for the acceleration of this system and then solve for the frequency (or rev/min) of the system?
Thanks again for the help.

Once you know the two tensions, you can calculate the total of the components that make up the centripetal force. From that you can get acceleration and frequency if that is what you are after.