# Breakdown of Circular Motion

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#1
Hi,
Is my understanding right that when centripetal force is less than m*r*omega^2, the particle takes off at a tangent where as when the centripetal force is greater than m*r*omega^2, the particle moves in projectile motion. I am talking in the context of a string strictly, not a rod.
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1 year ago
#2
(Original post by Tesla3)
Hi,
Is my understanding right that when centripetal force is less than m*r*omega^2, the particle takes off at a tangent where as when the centripetal force is greater than m*r*omega^2, the particle moves in projectile motion. I am talking in the context of a string strictly, not a rod.
what is generating the force, the tension in the string?
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#3
(Original post by mqb2766)
what is generating the force, the tension in the string?
Tension and at some points components of weight are effecting the motion as well.
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1 year ago
#4
(Original post by Tesla3)
Tension and at some points components of weight are effecting the motion as well.
Ok, can you be a bit more specific. In the horizontal plane so no gravity etc or .... It helps tp give a clear answer.
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#5
(Original post by mqb2766)
Ok, can you be a bit more specific. In the horizontal plane so no gravity etc or .... It helps tp give a clear answer.
Well, its a vertical circle so gravity is playing some role.............
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1 year ago
#6
(Original post by Tesla3)
Well, its a vertical circle so gravity is playing some role.............
It does, but only if you say its vertical motion. So what is the force equation perpendicular to the tangent (along the string)? If you have circular motion, the tension is the difference between centripetal and gravity. If the string snaps, there is a parabolic motion of a projectile under gravity. if the tension goes to zero (and then negative so you don't have circular motion), its again parabolic as the string plays no role hence the circular/centripetal effect does not apply.
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#7
(Original post by mqb2766)
It does, but only if you say its vertical motion. So what is the force equation perpendicular to the tangent (along the string)? If you have circular motion, the tension is the difference between centripetal and gravity. If the string snaps, there is a parabolic motion of a projectile under gravity. if the tension goes to zero (and then negative so you don't have circular motion), its again parabolic as the string plays no role hence the circular/centripetal effect does not apply.
The force equation perpendicular to the tangent is :
m*g*sin(theta) = m*r*(d^2theta/dt^2)
Also what do you mean by the fact that tension is the difference between centripetal and gravity. Just don't get what you are saying....
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1 year ago
#8
(Original post by Tesla3)
The force equation perpendicular to the tangent is :
m*g*sin(theta) = m*r*(d^2theta/dt^2)
Also what do you mean by the fact that tension is the difference between centripetal and gravity. Just don't get what you are saying....
Where is the tension in that equatiion?
http://hyperphysics.phy-astr.gsu.edu...s/cirvert.html
Is pretty good.
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1 year ago
#9
Let's look at two examples: one horizontal and one vertical.

Take a mass, attatched to a string of length orbiting with angular frequency .

Horizontal:
We know when the string snaps the mass will move off tangentially.

Vertical:
Consider the forces acting on the mass at the top of the circle, where is the tension in the string.
If the string is to remain taught, thus able to deliver force, we need .
If the string loses tension the mass will move as a parabola as the only force acting is gravity.
Now consider the forces acting on the mass at the bottom of the circle, so .
If this is greater than the maximium tension the string can tolerate, the mass will undergo parabolic motion as the only force acting is gravity.
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#10
(Original post by mqb2766)
Where is the tension in that equatiion?
http://hyperphysics.phy-astr.gsu.edu...s/cirvert.html
Is pretty good.
oh sorry the force equation perpendicular to tangent is:
T - mgcos(theta) = m*r*omega^2

and the force equation parallel to tangent is:
mgsin(theta) = m*r*(d^2theta/dt^2)

Also when you say that " If you have circular motion, the tension is the difference between centripetal and gravity. " Do you mean to say motion under gravity and motion under centripetal force (i.e. circular motion).
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#11
(Original post by anon2.718)
Let's look at two examples: one horizontal and one vertical.

Take a mass, attatched to a string of length orbiting with angular frequency .

Horizontal:
We know when the string snaps the mass will move off tangentially.

Vertical:
Consider the forces acting on the mass at the top of the circle, where is the tension in the string.
If the string is to remain taught, thus able to deliver force, we need .
If the string loses tension the mass will move as a parabola as the only force acting is gravity.
Now consider the forces acting on the mass at the bottom of the circle, so .
If this is greater than the maximium tension the string can tolerate, the mass will undergo parabolic motion as the only force acting is gravity.
In vertical motion, wouldn't the force be T - mgsin(theta) = m*r*omega^2 ?
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1 year ago
#12
(Original post by Tesla3)
oh sorry the force equation perpendicular to tangent is:
T - mgcos(theta) = m*r*omega^2

and the force equation parallel to tangent is:
mgsin(theta) = m*r*(d^2theta/dt^2)

Also when you say that " If you have circular motion, the tension is the difference between centripetal and gravity. " Do you mean to say motion under gravity and motion under centripetal force (i.e. circular motion).
Tension is the sum / difference of centripetal and resolved gravity. If its too large the string breaks. If it goes to zero, the string goes slack. In both cases you then have parabolic (quadratic - suvat) motion.
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1 year ago
#13
(Original post by Tesla3)
In vertical motion, wouldn't the force be T - mgsin(theta) = m*r*omega^2 ?
Please define theta before throwing it out lol
And I said to consider at the top and bottom of the circle.
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#14
(Original post by mqb2766)
Tension is the sum / difference of centripetal and resolved gravity. If its too large the string breaks. If it goes to zero, the string goes slack. In both cases you then have parabolic (quadratic - suvat) motion.
I thought that when the string breaks in vertical motion, i.e. when centripetal acceleration is less than m*r*omega^2, the particle flies of at a tangent.
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1 year ago
#15
(Original post by Tesla3)
I thought that when the string breaks in vertical motion, i.e. when centripetal acceleration is less than m*r*omega^2, the particle flies of at a tangent.
You have to consider when T<=0 or the string breaks.When positive (and not too large), the strings tension is equal to the sum / difference of centripetal & gravity and circular motion will happen.
The particle does fly off at a tangent, but this is the initial condition for parabolic motion.
Last edited by mqb2766; 1 year ago
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#16
(Original post by mqb2766)
You have to consider when T<=0 or the string breaks.When positive (and not too large), the strings tension is equal to the sum / difference of centripetal & gravity and circular motion will happen.
The particle does fly off at a tangent, but this is the initial condition for parabolic motion.
what happens when the string breaks?
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1 year ago
#17
(Original post by Tesla3)
what happens when the string breaks?
Parabolic or suvat motion under gravity.
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#18
(Original post by anon2.718)
Please define theta before throwing it out lol
And I said to consider at the top and bottom of the circle.
Oh sorry didn't read the top and bottom word accidentally,..............

Btw when you say that " If the string is to remain taught, thus able to deliver force, we need T > 0 (i.e. m*r*omega^2 - mg > 0 " .

Shouldn't it be T > 0 (i.e. m*r*omega^2 - mgcos(theta) > 0 ) ?
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#19
(Original post by mqb2766)
Parabolic or suvat motion under gravity.
So you are saying that in a vertical circle, if the string breaks, the particle either moves in parabolic or suvat motion. What decides that it is suvat or parabolic?
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1 year ago
#20
(Original post by Tesla3)
Shouldn't it be T > 0 (i.e. m*r*omega^2 - mgcos(theta) > 0 ) ?
What is cos(0)? Similarly at the bottom of the circle we do the same thing, what is cos(pi)?
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