Circular motion: Finding tension and theta

In summary, to estimate the force a person must exert to make a 0.200 kg ball revolve in a horizontal circle of radius 0.600 m, first find the acceleration by converting revolutions per second into m/s. Then use the equation Fnet=ma to solve for the net force. Next, divide the weight of the ball by the net force to find the angle θ. Finally, use the equations mg=FT*sinθ and m*v^2/r=FT*cosθ to solve for the magnitude of FT and the angle θ.
  • #1
swede5670
78
0

Homework Statement


Estimate the force a person must exert on a string attached to a 0.200 kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 1.40 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle θ it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?]

Homework Equations


Fnet=ma
w=v/2(pi)(r)
so
v=w*2(pi)(r)

The Attempt at a Solution


I've made a free body diagram but I was not shown how to solve any problems like this so I'm a little unsure as to how I'm going to begin. I decided to break up tension into horizontal and vertical components
horizontal tension = cos (theta) * t
vertical tension = sin (theta) * t
Then I found the acceleration by converting the revolutions per second into m/s and then finally finding the acceleration which is 1.4*(2*pi*.6)=5.277 then i converted that into acceleration since fnet=ma. v^2/r=a so 27.84/.600=46.41m/s/s
I then take this acceleration and multiply it by mass which is .2kg and I get fnet=9.28.
And here's where I become confused. I'm trying to solve for tension so I can't get rid of that variable but I'm fairly sure I need to substitute something in here for the equation to work.
T*sine(theta)=9.28

Since the ball is not moving up or down we can say that
t*cosine(theta)-mg=0
which ends up as
t*cos(theta)=1.962

I have 2 variables (tension and theta) and I'm not sure where/how to substitute for this problem.
 
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  • #2
m*v^2/r = FT*cos theta...(1)
mg=FT*sin theta...(2)
Divide (2) by (1) and find the angle.
 
  • #3
I'm not sure how to do that, I understand both of the equations but how do you divide one equation by the other? Are you telling me to substitute?
And are you sure about your two equations? I have something similar in my first post but the sin and cos are switched
you said: m*v^2/r = FT*cos theta but I thought it would be m*v^2/r = FT*sin
The reason I thought it was sin here was because I'm using sine to find the horizontal component of tension, cos is the vertical component isn't it?
And on the second equation mg=FT*sin theta I said it was mg=ft*cos (theta) because mg must equal the vertical component not the horizontal one.
 
  • #4
^^^^^ Bump

I really need this for my quiz review packet, please help!
 
  • #5
swede5670 said:
I'm not sure how to do that, I understand both of the equations but how do you divide one equation by the other?

Since they are equations it's like dividing each side by the same thing isn't it?

This will yield the angle θ as the tangent or cotangent of θ depending on which way you do it.

Armed with the angle can't you determine the FT?
 
  • #6
Since they are equations it's like dividing each side by the same thing isn't it?

This will yield the angle θ as the tangent or cotangent of θ depending on which way you do it.

Armed with the angle can't you determine the FT?

9.28= FT*sin (theta)

1.96200=ft*cos (theta)

so Ft cancels and i divide 9.28/1.962=4.72986748
and then tan (4.72986748) to get theta?

The problem is that only gives me .0827 and i have a feeling that is not what my angle should be.
 
  • #7
Wait, I have to use tan^-1 right? if that's the case I get 78.06 degrees which seems a lot more reasonable. However, I checked and the answer is wrong, does anyone know what I did incorrectly?
 
  • #8
That is right.
 
  • #9
Well the website I had to submit it to said it was wrong, so maybe it's wrong?
 
  • #10
swede5670 said:
9.28= FT*sin (theta)

1.96200=ft*cos (theta)
Where did you get those numbers?
 
  • #11
I was trying to find acceleration when I found 9.28
1.4*(2*pi*.6)=5.277 m/s (velocity)
v^2/r=a so 27.84/.600=46.41m/s/s (acceleration)
and I found it when I put it in the Fnet=ma equation
I mutliplied mass by the acceleration I had found (46.41*.20) which equals 9.28

I found 1.962 as mg, the mass was .2 kg and multiplied it by the acceleration of gravity which is 1.962
 
  • #12
You should be using the equations provided by rl.bhat in post #2.
 

1. What is circular motion?

Circular motion refers to the movement of an object in a circular path around a fixed point. It can be described as the combination of linear motion and centripetal acceleration.

2. How do you find tension in circular motion?

To find tension in circular motion, you need to use Newton's second law of motion, which states that the sum of all forces acting on an object is equal to its mass multiplied by its acceleration. In circular motion, tension is one of the forces acting on the object, so you can use this formula to solve for tension.

3. What is the role of tension in circular motion?

Tension is the force that acts on an object to keep it moving in a circular path. In circular motion, tension is always directed towards the center of the circle and is responsible for providing the necessary centripetal force to maintain the circular motion.

4. How do you find theta in circular motion?

To find theta (θ) in circular motion, you need to use trigonometry. Theta represents the angle between the radius of the circle and the tension force acting on the object. You can use the inverse trigonometric functions (sin⁻¹, cos⁻¹, tan⁻¹) to find theta depending on the given information in the problem.

5. What are the units for tension in circular motion?

The units for tension in circular motion are Newtons (N), which is the unit for force in the metric system. It can also be expressed in other units such as pounds (lb) in the imperial system. Tension is a vector quantity, so it has both magnitude and direction, and its units reflect that as well.

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