Rotational Force Problem

In summary, The problem involves a conical pendulum with a 89.0 kg bob on a 10.0 m wire making an angle of 5.00° with the vertical. The equations for horizontal and vertical tension are Tcos(\theta) - mg = 0 and Tsin(\theta) - mg = 0, respectively. The acceleration can be found using the formula a_r = gtan(\theta).
  • #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 a lot 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.
 
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  • #2
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
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
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 a lot of trouble on
 
  • #5
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]
 

1. What is rotational force?

Rotational force, also known as torque, is the measure of the force applied to an object to cause it to rotate around an axis.

2. How is rotational force different from linear force?

Rotational force and linear force are both types of force, but they act on objects in different ways. Linear force causes an object to move in a straight line, while rotational force causes an object to rotate around an axis.

3. What factors affect the magnitude of rotational force?

The magnitude of rotational force depends on the amount of force applied and the distance between the axis of rotation and the point where the force is applied. This can also be affected by the mass and shape of the object.

4. How is rotational force calculated?

Rotational force is calculated by multiplying the force applied by the distance from the axis of rotation to the point where the force is applied. The formula for rotational force is F = r x F, where F is force in newtons, r is distance in meters, and F is torque in newton-meters.

5. What are some real-life examples of rotational force?

Some examples of rotational force in everyday life include opening a door, using a wrench to tighten a bolt, and spinning a top. In science, rotational force is also important in understanding the movement of planets and other celestial bodies in space.

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