Find center of mass using rotational equilibrium

In summary, the problem involves a nonuniform bar suspended at rest by two massless cords. Using rotational equilibrium, the distance x from the left end of the bar to its center of mass can be calculated by setting up conditions for no vertical, horizontal, and rotational motion. Solving for x using these conditions and given values, we get x = 2.13 m. However, a mistake was made in the calculation of T1, resulting in an incorrect value for x. The correct value is found by taking the ratio of equations 1 and 2, giving x = 0.1453 m.
  • #1
farleyknight
146
0

Homework Statement



A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown in Fig. 12-45. One cord makes the angle θ = 15.5° with the vertical; the other makes the angle φ = 58.5° with the vertical. If the length L of the bar is 6.08 m, compute the distance x from the left end of the bar to its center of mass.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c12/fig12_39.gif

Homework Equations



[tex]\tau_{net} = 0[/tex]

The Attempt at a Solution



Okay, so this chapter is mainly on static equilibrium, and so we're going to use the rotational equilibrium to determine the value of x in this problem.

There are exactly three forces acting on the bar: The force from the left cord, the force from the right cord, (both equal in magnitude), and the force of gravity. If we make the center of coordinate system the same as the center of gravity (unknown, I realize, but we can still use x), then the y-coordinate of the force of gravity falls out of our equation, and we can take the torques as:

[tex]\tau_{net} = T \sin( \bar{\theta} )(-x) + T \sin( \bar{\phi} )(L - x)[/tex]

where I've taken [tex]\bar{\theta} = 90 - \theta[/tex], the right triangle compliment of [tex]\theta[/tex]

Solving for x, we have

[tex]T \sin( \bar{\theta} ) x + T \sin( \bar{\phi} ) x = T \sin( \bar{\phi} ) L [/tex]

Pulling out T and moving our sines to the right

[tex]x = \frac{ \sin( \bar{\phi} ) L }{ \sin( \bar{\phi} ) + \sin( \bar{\theta} ) }[/tex]

Then plugging in values

[tex]x = \frac{ \sin(31.5) (6.08) }{ \sin(31.5) + \sin(74.5) } = 2.13 m[/tex]

However, this isn't correct. I imagine my mistake is probably a simple one.. since I managed to do about half the homework problems in this chapter without any real need for help. But that was about a week ago or so, and it seems the ideas flew out of my head :) So any help would be grateful.
 
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  • #2
In this problem T1 and T2 are not the same.
Since the rod is in equilibrium, there is no vertical motion, horizontal motion and the rotation of the rod.
Write the conditions for these three cases. Then solve for x.
 
  • #3
rl.bhat said:
In this problem T1 and T2 are not the same.
Since the rod is in equilibrium, there is no vertical motion, horizontal motion and the rotation of the rod.
Write the conditions for these three cases. Then solve for x.

Okay, so then we would have

[tex]F_{net,x} = - T_1 \cos(\bar{ \theta }) + T_2 \cos( \bar{ \theta } )[/tex]

And also

[tex]F_{net,y} = -mg + T_1 \sin(\bar{ \theta }) + T_2 \sin(\bar{ \theta })[/tex]

We create a relation for [tex]T_1[/tex] and [tex]T_2[/tex]:

[tex]T_1 = T_2 \frac{ \cos(\bar{\phi}) }{ \cos(\bar{\theta}) }[/tex]

We then use this relation to solave for [tex]T_1[/tex] in the second equation

[tex] mg = T_1 \sin( \bar{ \theta } ) + T_1 \frac{ \cos(\bar{\theta}) }{ \cos(\bar{\phi}) } \sin(\bar{\phi})[/tex]

or more simply

[tex]mg = T_1 \sin(\bar{\theta}) + T_1 \cos( \bar{\theta} ) \tan( \bar{ \phi } )[/tex]We can then find [tex]T_1[/tex] and [tex]T_2[/tex] as values of m and g

[tex] T_1 = \frac{ mg }{ \sin(\bar{\theta}) + \cos(\bar{\theta})\tan(\bar{\phi}) } [/tex]

or [tex] T_1 = 11.425 m [/tex]

and

[tex] T_2 = \frac{ mg \cos(\bar{\phi}) }{ \cos(\bar{\phi}) ( \sin(\bar{\theta}) + \cos(\bar{\theta})\tan(\bar{\phi})) } [/tex]

or [tex] T_2 = 3.581 m [/tex]

We then plug these into the equation for torque

[tex]\tau_{net} = 0 = T_1 \sin(\bar{\theta})(-x) + T_2 \sin(\bar{\phi})(L - x)[/tex]

Solving for x we have

[tex]x = \frac{ T_2 \sin(\bar{\phi}) }{ T_1 \sin(\bar{\theta}) + T_2 \sin(\bar{\phi}) }[/tex]

or

[tex]x = \frac{ 3.581 \sin(\bar{\phi}) }{ 11.425 \sin(\bar{\theta}) + 3.581 \sin(\bar{\phi}) }[/tex]

But evaluating this gets me a pretty small number, namely [tex]x = 0.1453[/tex], which isn't correct.
 
Last edited:
  • #4
Your T1 calculation is wrong. You can skip this step.
From F(netx) you get
T1*cos(90 - θ) = T2*cos(90 - φ) ------(1)
Using Τ(net) you get
T1*sin(90 - θ)*x = T2*sin(90-φ)*( L - x )---(2)
Take the ratio of eq 1 and 2, and solve for x.
 

FAQ: Find center of mass using rotational equilibrium

What is rotational equilibrium?

Rotational equilibrium is a state in which an object is not rotating or is rotating at a constant rate. This means that the net torque acting on the object is zero.

How is the center of mass related to rotational equilibrium?

The center of mass is the point at which an object's mass is evenly distributed. In rotational equilibrium, the center of mass is also the point around which the object rotates without any external torque.

What is the formula for finding the center of mass using rotational equilibrium?

The formula for finding the center of mass using rotational equilibrium is: xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn) where xcm is the position of the center of mass and m1, m2, ... mn are the masses of individual parts of the object and x1, x2, ... xn are their respective distances from a chosen reference point.

What are the units for the center of mass?

The units for the center of mass depend on the units used for the object's mass and position. In the formula above, if the masses are in kilograms and the distances are in meters, then the center of mass will be in meters.

What are some real-life applications of finding the center of mass using rotational equilibrium?

Finding the center of mass using rotational equilibrium is important in various fields such as engineering, physics, and sports. It is used in designing stable structures, analyzing the stability of vehicles, and determining the optimal body position for athletes in sports like diving and gymnastics.

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