Circular Motion of ball tension

AI Thread Summary
The discussion revolves around calculating the tension in two cords supporting a 1.75 kg ball being whirled in a circular motion with a radius of 95 cm. The user initially attempts to solve the problem using the equations of motion but encounters errors in sign conventions when resolving the forces. Corrections are suggested, emphasizing the importance of proper vector component analysis for both vertical and horizontal forces. The final calculations yield tensions of approximately 1650.72 N and 1630.92 N, which are deemed unreasonably high, prompting further reflection on the physical implications of such forces. The conversation highlights the need for careful attention to detail in physics problem-solving.
Nytrox
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Homework Statement


In the device shown below, a 1.75 kg ball suspended by two cords is whirled around in a circle of radius 95 cm. The ball makes 10 revolutions in 2 seconds. Find the tension in each of the two cords (remember: the tension in each cord acts toward the pole).
https://www.physicsforums.com/attachments/29633
Well...the attachment isn't quite working...so I'll attempt to draw the picture...

|..\ <--This angle 30 degrees
|...\ <--Cord 1
|...\
|...O <- Ball.....Radius from ball to pole = .95 meters (95 cm)
|.../......Angle of cords are both 30 degrees from the pole.
|.../ <--Cord 2.....Mass of ball is 1.75 kg
|.../
|.../ <--This angle 30 degrees
|./
^
Pole


Homework Equations


Fc = mV^2/r
Vt=2pi r/T
Fg=mg
F=ma
Ftx=Ftsin theta
Fty=Ftcos theta


The Attempt at a Solution



r=.95m
T=.2 seconds

Fty1-Fty2=Fg
Ft1cos30 - Ft2cos210 = 1.75kg x 9.8 m/s^2
Ft1+Ft2=19.8

Ftx1+Ftx2=Fc
Ft1sin30 + Ft2sin210 = m((2 pi r)/T)^2 / r
.5Ft1 - .5Ft2 = 1.75kg x (29.85 m/s)^2 / .95m
.5Ft1 - .5Ft2 = 1640.82N
Ft1 - Ft2 = 3281.64N

Ft1 = 19.8 - Ft2
19.8 - 2Ft2 = 3281.64N
Ft2 = -1630.92N
Ft1 = 1650.72N

I'm not sure if what I did is completely right. Please help. Thanks!
 
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Nytrox said:
Fty1-Fty2=Fg
Ft1cos30 - Ft2cos210 = 1.75kg x 9.8 m/s^2
Ft1+Ft2=19.8
Right there...
That's not right. First of all, you get a negative quantity when you cos210, but that does't mean that you add the two components. You ended up doing a double negative there. Fty1 pulls up so it's positive, Fty2 pulls down so it's negative.

You did the same thing in the next part. Both horizontal components are pulling in the same direction, so they should add together to become the net centripetal force here.
 
Nytrox said:

Homework Statement


In the device shown below, a 1.75 kg ball suspended by two cords is whirled around in a circle of radius 95 cm. The ball makes 10 revolutions in 2 seconds. Find the tension in each of the two cords (remember: the tension in each cord acts toward the pole).
https://www.physicsforums.com/attachments/29633
Well...the attachment isn't quite working...so I'll attempt to draw the picture...

|..\ <--This angle 30 degrees
|...\ <--Cord 1
|...\
|...O <- Ball.....Radius from ball to pole = .95 meters (95 cm)
|.../......Angle of cords are both 30 degrees from the pole.
|.../ <--Cord 2.....Mass of ball is 1.75 kg
|.../
|.../ <--This angle 30 degrees
|./
^
Pole


Homework Equations


Fc = mV^2/r
Vt=2pi r/T
Fg=mg
F=ma
Ftx=Ftsin theta
Fty=Ftcos theta


The Attempt at a Solution



r=.95m
T=.2 seconds

Fty1-Fty2=Fg
Ft1cos30 - Ft2cos210 = 1.75kg x 9.8 m/s^2
Ft1+Ft2=19.8
You have a signage error. That last term should be -Ft2cos30
Ftx1+Ftx2=Fc
Ft1sin30 + Ft2sin210 = m((2 pi r)/T)^2 / r
same error in signs...last term should be +F2sin30
.5Ft1 - .5Ft2 = 1.75kg x (29.85 m/s)^2 / .95m
.5Ft1 - .5Ft2 = 1640.82N
Ft1 - Ft2 = 3281.64N

Ft1 = 19.8 - Ft2
19.8 - 2Ft2 = 3281.64N
Ft2 = -1630.92N
Ft1 = 1650.72N

I'm not sure if what I did is completely right. Please help. Thanks!
you were on track, but slipped on the plus and minus signs. When determining vector components, it is often best to draw a sketch using right triangles to determine the trig relationships.
 
ok. So it should be...

Fty1-Fty2=Fg
Ft1cos30 - Ft2cos30 = 1.75kg x 9.8 m/s^2
Ft1-Ft2=19.8

Ftx1+Ftx2=Fc
Ft1sin30 + Ft2sin30 = m((2 pi r)/T)^2 / r
.5Ft1 + .5Ft2 = 1.75kg x (29.85 m/s)^2 / .95m
.5Ft1 + .5Ft2 = 1640.82N
Ft1 + Ft2 = 3281.64N

Ft1 = 19.8 + Ft2
19.8 + 2Ft2 = 3281.64N
Ft2 = 1630.92N
Ft1 = 1650.72N

Is this the right answer? It seem unreasonably large...
 
Nytrox said:
ok. So it should be...

Fty1-Fty2=Fg
Ft1cos30 - Ft2cos30 = 1.75kg x 9.8 m/s^2
Ft1-Ft2=19.8

Ftx1+Ftx2=Fc
Ft1sin30 + Ft2sin30 = m((2 pi r)/T)^2 / r
.5Ft1 + .5Ft2 = 1.75kg x (29.85 m/s)^2 / .95m
.5Ft1 + .5Ft2 = 1640.82N
Ft1 + Ft2 = 3281.64N

Ft1 = 19.8 + Ft2
19.8 + 2Ft2 = 3281.64N
Ft2 = 1630.92N
Ft1 = 1650.72N

Is this the right answer? It seem unreasonably large...

Does it?

1600 N, is that a lot? (It's the weight of how many people?)

30 m/s, is that fast? (how many mph is that?

This is a 4 lb object twirling in a 1 yard radius circle, at a pretty fast speed. How much force do you think is appropriate?
 
well 30 m/s is around 67 miles per hour. That's really fast.
and 1600 Newtons is around 360 pounds of force
 

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