How to find the tension of the cord (Conical Pendulum)?

In summary, the conversation is about finding the tension of a cord attached to a mechanical bat flying in a circle on the ceiling, given the mass of the bat, the time for one revolution, the length of the cord, and the height from the ceiling. The suggested equation is Tension*cos(x) = mg, where x is the angle between the cord and the vertical direction. The value of x is determined to be 45 degrees, and the final step is to plug in the values to solve for the tension, which is calculated to be 1.86 N. The importance of understanding the equation and identifying the angle correctly is emphasized.
  • #1
onhcetum
2
0

Homework Statement



Hey, we have this mechanical bat that is attached to a cord and its flying around in a circle on the ceiling. Here is all the information that I have gathered.

Mass of the bat: 0.1345 kg
1.609 seconds per revolution
length of cord: 0.92 m
height from ceiling: 0.65 m

Here is what it looks like if it helps. I need to find the tension of the cord.

http://i49.tinypic.com/2ujkyme.jpg

Steps and equations would be very helpful.

Thank you!


Homework Equations



I don't know what equation to use or where to start. I think it's...

Tension*cos(x) = mg

The Attempt at a Solution



I don't know where to start; I've drawn a picture if you might help anyone.
 
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  • #2
onhcetum said:
I don't know what equation to use or where to start. I think it's...

Tension*cos(x) = mg
That actually might be the right equation, depending on what x is. But you shouldn't just use it without understanding where it comes from. So: (1) how did you come up with that equation? (2) which angle is x?
 
  • #3
cos^(-1) (0.65/0.92) = ~ 45 degrees

It is 45 degrees.

I think I read it somewhere.

What would be my next step? Just plugging everything in?

Tension = (9.8m/s^2)(0.1345 kg)/(cos 45) = 1.86 N?
 
  • #4
I didn't ask what the value of x is, I asked which angle it is, i.e. identify where on the diagram it is. But the fact that you wrote "cos^(-1) (0.65/0.92)" suggests that you identified it correctly. So that works.

But for your own sake, I'd suggest continuing to think about this until you understand why that's the right answer. "I think I read it somewhere" doesn't really help you get anything out of the problem. (Sometimes it's the best you can do, but this is not one of those times)
 

Related to How to find the tension of the cord (Conical Pendulum)?

1. What is a conical pendulum?

A conical pendulum is a type of pendulum that moves in a conical motion instead of a simple back and forth motion. It consists of a mass attached to a cord or string that is suspended from a fixed point and swings in a circular path.

2. How do I find the tension of the cord in a conical pendulum?

To find the tension of the cord in a conical pendulum, you will need to know the mass of the object, the radius of the circular path, and the speed of the object. The formula for tension is T = (m*v^2)/r, where T is tension, m is mass, v is speed, and r is radius.

3. What factors affect the tension of the cord in a conical pendulum?

The tension of the cord in a conical pendulum is affected by the mass of the object, the radius of the circular path, and the speed of the object. Additionally, the length and elasticity of the cord can also affect the tension.

4. How can I measure the tension of the cord in a conical pendulum?

You can measure the tension of the cord in a conical pendulum by using a spring scale or a force sensor. Attach the scale or sensor to the cord and measure the force reading. This will give you the tension of the cord at that particular point.

5. Why is it important to know the tension of the cord in a conical pendulum?

Knowing the tension of the cord in a conical pendulum is important because it affects the motion of the pendulum. The tension determines the speed and radius of the circular motion, and can also affect the period of the pendulum. Understanding the tension can help in predicting the behavior of the pendulum and making adjustments for more accurate results.

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