angular frequency


by tony873004
Tags: angular, frequency
tony873004
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Mar4-05, 12:08 AM
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16. A cord can support a maximum tension of 1.0 * 10^4N before it breaks. The cord is used to swing a 15-kg stone in a circular path of radius 1.0 m over an icy surface. What cord tension is necessary as a function of angular frequency? What is the maximum angular frequency such that the cord will not break?

Aside from this question, the book doesn't mention the term "angular frequency" for another 200 pages. And it describes it as something that doesn't even relate to this problem. And it calls it ω which is what the current chapter uses for angular velocity.

I believe this problem wants me to solve either:

What is the max angular speed without breaking the cord, and what tension is necessary as a function of angular speed
(ie: Tension(ω) = .....)

or

What is the maximum amount of rotations per second will the stone can make without breaking the cord, and what tension is necessary as a function of rotations per second

(ie: Tension(rps) = ....)

I'm not asking for help in doing the problem. I just want to know what they're asking

Also, what does the icy surface have to do with it? It does not give the mass of the person swining the stone, so I can't compute a barycenter. I guess in writing the formula for the Tension function, the person's mass could be introduced as a variable. But that's far beyond the scope of this chapter. There is nothing in the chapter on circular motion that talks about an origin that is not fixed. And there's no guarantee that the person isn't standing on rock surrounded by ice in which case the stone would still fulfill the question's requirements of being swung over an icy surface.


thanks...
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xanthym
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Mar4-05, 01:49 AM
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Quote Quote by tony873004
16. A cord can support a maximum tension of 1.0 * 10^4N before it breaks. The cord is used to swing a 15-kg stone in a circular path of radius 1.0 m over an icy surface. What cord tension is necessary as a function of angular frequency? What is the maximum angular frequency such that the cord will not break?
Standard definitions are:
{Angular Frequency} = {Angular Velocity} = ω = {Linear Velocity}/{Circle Radius} = v/r
{Frequency} = f = ω/(2π)

{Centripetal Force Required} = (mv2)/r =
= m*r*(v/r)2 =
= m*r*ω2

{Cord Tension Required} = m*r*ω2

Set above equation equal to {(Max Cord Tension)=(1.0e4 N)} and solve for Max Angular Frequency "ωmax" using problem values of {m=(15 kg)} and {r=(1 m)}:

(1.0e4 N) = (15 kg)*(1 m)*(ωmax)2
{Max Angular Frequency} = ωmax = (25.82 radians/sec)


~~
cepheid
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Mar4-05, 02:00 AM
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your second definition is not correct. Angular frequency is also [itex] \omega = 2\pi f [/itex]. It is 2 pi times the frequency f. That's why it's called the angular frequency. Note that when the motion is actually rotational, the angular velocity and angular frequency are exactly the same thing. For other non-rotational periodic motion (e.g. SHM), it makes no sense to speak of angular velocity, since nothing is actually rotating, but we can still speak of the angular frequency 2pi*f as 2 pi times the frequency of the oscillations. It can be thought of as the analogous angular velocity that a body undergoing uniform circular motion would have if the motion had that same frequency f.

ehild
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Mar4-05, 02:12 AM
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angular frequency


Quote Quote by tony873004
16. A cord can support a maximum tension of 1.0 * 10^4N before it breaks. The cord is used to swing a 15-kg stone in a circular path of radius 1.0 m over an icy surface. What cord tension is necessary as a function of angular frequency? What is the maximum angular frequency such that the cord will not break?

Aside from this question, the book doesn't mention the term "angular frequency" for another 200 pages. And it describes it as something that doesn't even relate to this problem. And it calls it ω which is what the current chapter uses for angular velocity.
Angular frequency is just 2pi times the frequency.

[tex] \omega = 2\pi f[/tex]

And it happens to be the same as the magnitude of the angular velocity. (The angular velocity is a vector, parallel to the axis of rotation and its direction is so that the rotation corresponds to the motion of a right screw.) Its magnitude -the angular speed- is the angle "travelled' in unit time. As the angle is 2pi during one revolution and the time T needed for it is the time period, the angular speed is 2pi/T in case of uniform circular motion. But it is just a kind of periodic motion, and that has got a frequency, reciprocal of the time period: f=1/T. So the angular speed of the uniform circular motion is

[tex] \omega = \frac{d\alpha}{d t} = 2\pi/T= 2\pi f [/tex]


I believe this problem wants me to solve either:

What is the max angular speed without breaking the cord, and what tension is necessary as a function of angular speed
(ie: Tension(ω) = .....)

or

What is the maximum amount of rotations per second will the stone can make without breaking the cord, and what tension is necessary as a function of rotations per second

(ie: Tension(rps) = ....)
You have to give the tension as function of omega, and you can call it either angular frequency or angular speed, and also you have to give the maximum of omega, which is 2pi times the revolutions per second.



Also, what does the icy surface have to do with it? It does not give the mass of the person swining the stone, so I can't compute a barycenter. I guess in writing the formula fr the Tension function, the person's mass could be introduced as a variable. But that's far beyond the scope of this chapter. There is nothing in the chapter on circular motion that talks about an origin that is not fixed. And there's no guarantee that the person isn't standing on rock surrounded by ice in which case the stone would still fulfill the question's requirements of being swung over an icy surface.
Just assume that the center of the circle is fixed. The icy surface means that friction should be ignored, the only horizontal force on the stone is from the cord.

Now you have to know about the centripetal force. This force is needed to maintain circular motion along a circle of radius R and with speed v or angular speed omega=v/R. And this force should be equal to the tension in the rope.

[tex] T = F_{cp}= \frac{v^2}{R}=R\omega^2[/tex].

ehild
xanthym
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Mar4-05, 02:29 AM
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Quote Quote by ehild


[tex] T = F_{cp}= \frac{v^2}{R}=R\omega^2[/tex].

ehild
The last equation in the previous msg should also contain the Mass "M" of the object:

[tex] T = F_{cp}= \color{red}M\color{black}\frac{v^2}{R}=\color{red}M\color{black}R\omega ^2[/tex]



~~
ehild
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Mar4-05, 08:50 AM
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Quote Quote by xanthym
The last equation in the previous msg should also contain the Mass "M" of the object:

[tex] T = F_{cp}= \color{red}M\color{black}\frac{v^2}{R}=\color{red}M\color{black}R\omega ^2[/tex]



~~
Of course... I just left it out. Thanks.
tony873004
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Mar10-05, 10:25 PM
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Thanks for all your help on this problem last week. I was working on this a week before it was due, but now it is due tommorow. Reading your replies, I came up with:

[tex]T(\omega) = m\omega^2r[/tex] for the first part,

and after cancelling my units I came up with
[tex]\omega_{max}=25.8 /s[/tex]
Does /s equal Hertz? Should I be expressing this answer as
[tex]\omega_{max}=25.8 Hz[/tex]
Xerxes1986
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Mar10-05, 10:29 PM
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Quote Quote by tony873004
Thanks for all your help on this problem last week. I was working on this a week before it was due, but now it is due tommorow. Reading your replies, I came up with:

[tex]T(\omega) = m\omega^2r[/tex] for the first part,

and after cancelling my units I came up with
[tex]\omega_{max}=25.8 /s[/tex]
Does /s equal Hertz? Should I be expressing this answer as
[tex]\omega_{max}=25.8 Hz[/tex]
yes a Hertz is defined as 1/s
xanthym
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Mar10-05, 10:34 PM
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Quote Quote by tony873004

and after cancelling my units I came up with
[tex]\omega_{max}=25.8 /s[/tex]
Does /s equal Hertz? Should I be expressing this answer as
[tex]\omega_{max}=25.8 Hz[/tex]
See msg #2 in this thread (last line in RED).
tony873004
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Mar10-05, 11:16 PM
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Quote Quote by xanthym
See msg #2 in this thread (last line in RED).
Thank you very much. I was going to put down Hz or revolutions per second. Now I can see where the radians came from in the formula and they should have followed me down to my answer when I did this.


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