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Rotational Force Problem

  • Thread starter Sheneron
  • Start date
  • #1
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[SOLVED] Rotational Force Problem

Homework Statement


Consider a conical pendulum with a 89.0 kg bob on a 10.0 m wire making an angle of = 5.00° with the vertical (Fig. P6.9).

http://img89.imageshack.us/my.php?image=p613qy9.gif


The Attempt at a Solution


I am having alot of trouble figuring this out. I have tried to set up force diagrams but I am getting confused. I know that the force is the y direction is just mg but to make it this way it would have to be
[tex]Tsin(\theta) - mgsin(\theta) = 0[/tex] and I don't understand why it would be mg multiplied by the sine of the angle.
Also, for the second part I don't understand why the answer is not just mg times the sin of the angle.
Can someone help me figure out this problem? Thanks.
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
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Hi Sheneron,

I'm not certain which direction you're trying to write that equation for. However, you're right that mg is completely vertical; also, the tension has both vertical and horizontal components.

For this problem, you'll also need the acceleration (magnitude and direction). What do you get?
 
  • #3
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For the horizontal tension the equation I get is
[tex]Tcos(\theta) - mgsin(\theta) = 0[/tex]
For vertical I get
[tex]Tsin(\theta) - mgsin(\theta) = 0[/tex]

I am pretty sure those have to be the answers, but I can't see where they came from. I don't understand how to set the force diagram up.
 
  • #4
360
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Now I am very confused... ma would not equal 0 would it? To find the acceleration you would first have to know the tension, which I am having alot of trouble on
 
  • #5
360
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ohhhhhhhh I think I got it

[tex]Tcos(\theta) - mg = 0[/tex]
[tex]T = \frac{mg}{cos(\theta)}[/tex]

[tex]T_y = Tcos(\theta) = \frac{mg}{cos(\theta)}cos(\theta) = mg[/tex]

[tex]T_x = Tsin(\theta) = \frac{mg}{cos(\theta)}sin(\theta)[/tex]

[tex] F_x = ma_r[/tex]

[tex]\frac{mg}{cos(\theta)}sin(\theta) = ma_r[/tex]

[tex]a_r = \frac{mgtan(\theta)}{m} [/tex]
 

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