Finding Radius of Circular Movement w/ Angle & Angular Velocity

In summary, in this problem, the radius of the circle does depend on the length of the cord and the length of the horizontal bar, but this dependence is encoded in the value of the angle \alpha. This is due to the physics of the problem, where \alpha is affected by changes in these quantities. Solving for R using \alpha and the given information is equivalent to solving for R using the explicit dependencies on the lengths of the cord and horizontal bar. However, this involves solving a quartic equation and does not give any additional information.
  • #1
Dell
590
0
in the following question,

http://62.90.118.184/Index.asp?CategoryID=318

i am given the following only:
angle of the cord = [tex]\alpha[/tex]
angular velocity = [tex]\omega[/tex]

and am asked to find the radius

what i did was :
Newtons 2nd on the radial and "y" axis

Fy=T*cos([tex]\alpha[/tex])-mg=0 (T being tension in the cord)
T=mg/cos([tex]\alpha[/tex])

Fr = T*sin([tex]\alpha[/tex]) = mar =m[tex]\omega[/tex]2R

T*sin([tex]\alpha[/tex]) =m[tex]\omega[/tex]2R

now i use the T i found (T=mg/cos([tex]\alpha[/tex])) and i get
mg*tg([tex]\alpha[/tex])=m[tex]\omega[/tex]2R

R=g*tg([tex]\alpha[/tex])/[tex]\omega[/tex]2

which according to the answer sheet is the correct answer, but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??
 
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  • #2
Dell said:
but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??

The radius does depend on those quantities, but their value is encoded in the value of [itex]\alpha[/itex]. If [itex]\omega[/itex] is fixed and we increase the length of the cord then [itex]\alpha[/itex] increases (and if we decrease the length of the cord it decreases). Similarly, for a fixed [itex]\omega[/itex], an increase in the length of the horizontal bar will increase [itex]\alpha[/itex]. The physics of what is happening (spinning a mass on a string) means that the value of [itex]\alpha[/itex] has a dependency on these quantities.

If you wanted, you could make this dependency explicit. From the geometry of the problem you can show that,

[tex]
\operatorname{tg}{\alpha}=\frac{R-a}{\sqrt{b^2-\left(R-a\right)^2}}
[/tex]

where [itex]a[/itex] is the length of the horizontal bar and [itex]b[/itex] is the length of the cord. If you substitute this into your answer then you can get an expression for [itex]R[/itex] that depends explicitly on [itex]a[/itex] and [itex]b[/itex] instead of [itex]\alpha[/itex]. However rearranging the answer involves solving a quartic equation for [itex]R[/itex] which is not easy.

As an interesting aside, note that understanding the physics doesn't give us any 'information for free'. If we are given [itex]\alpha[/itex] then to find [itex]R[/itex] geometrically we require two more pieces of information: the length of the horizontal bar and the length of the string. To find [itex]R[/itex] from [itex]\alpha[/itex] using instead the physics of the situation, we still require two more pieces of information (the speed of rotation [itex]\omega[/itex] and the downward acceleration the ball would experience if removed from the string, [itex]g[/itex]).
 
  • #3


I would like to point out that the radius of circular movement depends on the centripetal force and the angular velocity, not the length of the cord or the top horizontal bar. The length of the cord and the top horizontal bar may affect the angle and the tension in the cord, but they do not directly affect the radius. This is because the centripetal force is directed towards the center of the circle, and the angular velocity is the rate at which an object moves around that center. Therefore, as long as the angle and angular velocity remain constant, the radius will also remain constant. This can be seen in the equation R=g*tg(\alpha)/\omega2, where R is the radius, g is the acceleration due to gravity, alpha is the angle, and omega is the angular velocity. As you can see, none of these variables are affected by the length of the cord or the top horizontal bar. Therefore, the radius does not depend on these factors.
 

1. What is the formula for finding the radius of circular movement?

The formula for finding the radius of circular movement is r = v/ω, where r is the radius, v is the velocity, and ω is the angular velocity.

2. How do I calculate the radius if I only have the angle and angular velocity?

If you only have the angle and angular velocity, you can use the formula r = v/(ωsinθ), where θ is the angle and ω is the angular velocity.

3. Can the radius of circular movement be negative?

No, the radius of circular movement cannot be negative. It represents the distance from the center of the circle to the object moving in a circular path.

4. What units should I use for the velocity and angular velocity in the formula?

The units for the velocity and angular velocity should be consistent. For example, if the velocity is in meters per second, the angular velocity should be in radians per second.

5. How do I convert the angular velocity from revolutions per minute (RPM) to radians per second?

To convert from RPM to radians per second, you can use the formula ω = (2πn)/60, where ω is the angular velocity in radians per second and n is the RPM.

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