speed and circular motion

[ssh18]

In a vertical circular motion, when the object reaches the top of the circular path, are both the tension from the rope and the gravitational force pointing downward (toward the center?) Does that mean that speed is the highest at the top? I'm a bit confused because I thought speed decreased as the object rises to the top of the circular path??

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[BloodySurgeon]

I think others in this forum are more adequate than I am, but i'll take a stab at it. First, are you taking about a ball connected to a sting with tension? (an object going through a loop is similar but with normal force). Well, anyways... if an object is in uniform circular motion the speed is constant (although the velocity isn't) and therefore the acceleration is constant as well. And although at the peak there are two forces in the same direction as the acceleration, the tension will be much smaller than at the bottom (negliable if we are considering a uniform circular motion because the string will be extended long enough for no tension to exist.)

If you could be more specific on what your question is, I can be a bit of more help.

 

[BloodySurgeon]

In a vertical circular motion, when the object reaches the top of the circular path, are both the tension from the rope and the gravitational force pointing downward (toward the center?) Does that mean that speed is the highest at the top? I'm a bit confused because I thought speed decreased as the object rises to the top of the circular path??

You think ur confused, now? Try "How much wood would a woodchuck chuck
if a woodchuck could chuck wood?"

Answer: 7281 Butt cords

 

[TheBoondocks]

In a vertical circular motion, when the object reaches the top of the circular path, are both the tension from the rope and the gravitational force pointing downward (toward the center?) Does that mean that speed is the highest at the top? I'm a bit confused because I thought speed decreased as the object rises to the top of the circular path??

in uniform circular motion the acceleration is constant. you're right, at the top mg and Fn=tension both point down. however, they have to be equal to (mv^2)/r This is the centripetal force that keeps it accelerating in the circle. The speed stays the same, since it's uniform, but velocity doesn't. What does change is the tension. At the top, Fn=T =(mv^2)/r -mg. so tension is less. at the bottom, the tension is the greatest since mg is opposite tension. the velocity remains the same but tension doesn't.

Now, if you're not considering uniform motion then what i stated above isn't valid. In reality, vertical circular motion can't be constant because of tangental velocity. at the bottom there is less potential energy and at the top there is more potential energy. the tangental velocity acts to slow down the object as it goes to the top and speed it up as it moves towards t the bottom. You are right. here's a good site, scroll down, and you should get what you're looking for. http://cnx.org/content/m14090/latest/

 

[ssh18]

Thanks for the response. I'm still a bit confused though.

So, when we're talking about uniform circular motion (vertical):
1. Speed is constant
2. At the top, the tension on the rope is the lowest and at the bottom of the circle, the tension is the highest

When talking about non-uniform circular motion (vertical):
1. Speed isn't constant (highest at the bottom, lowest at the top). However, I don't see where the tangential acceleration is at the top. The Fgrav is pointing down so isn't there only one acceleration component (radial pointing towards the center)? If there is no tangential acceleration at the top, why is the speed the lowest at the top? Also, what is happening to the tension force at the top vs. at the bottom?

 

[majik1213]

To make things clear, let us assume that you're swinging a rock in a circle, the plane of which lies vertically.

In a vertical circular motion, when the object reaches the top of the circular path, are both the tension from the rope and the gravitational force pointing downward (toward the center?)

yes. The vectors both point downward on the object; at your hand (the center of the circle), the vector due to gravity points down and that due to tension up.

Does that mean that speed is the highest at the top?

Yes; but then again, the speed never changes; only the velocity does.

I'm a bit confused because I thought speed decreased as the object rises to the top of the circular path??

It depends. If you are swinging the rock in a circle, then you are providing the necessary tension to hold the rock at a constant speed (but not velocity). If you instantaneously affix your end of the rope to a fixed rod, then over time the rock will slow down until it comes to a rest at the bottom of the circle.

 

[majik1213]

Thanks for the response. I'm still a bit confused though.

So, when we're talking about uniform circular motion (vertical):
1. Speed is constant
2. At the top, the tension on the rope is the lowest and at the bottom of the circle, the tension is the highest

In uniform circular motion, either there can be no gravity, or we must assume that the tension varies in a way that perfectly cancels out the gravitational influence of gravity on the net centripetal force. The answer to 1 is yes, and to 2 is yes, provided there is gravity.

When talking about non-uniform circular motion (vertical):
1. Speed isn't constant (highest at the bottom, lowest at the top). However, I don't see where the tangential acceleration is at the top. The Fgrav is pointing down so isn't there only one acceleration component (radial pointing towards the center)? If there is no tangential acceleration at the top, why is the speed the lowest at the top? Also, what is happening to the tension force at the top vs. at the bottom?

correct; speed cannot remain constant, or else the motion becomes uniform by definition. At the top the tangential acceleration is pointing in the direction the object would travel if all the forces suddenly vanished. If you are, for example, watching the object rotate before you in a clockwise path, then at the top the tangential acceleration, if non-zero, points to the right.

In practice, you cannot make a rock on a rope go in uniform motion if you spin it vertically. To do so would require that you accelerate it tangentially at a max when it's halfway going up, and that you negatively accelerate it with equally maximal magnitude tangentially when it's halfway down. You must smoothly change this acceleration much as a cosine or sine wave. If you do that, then props.

 

[RSAgator]

Is non-uniform circular motion required knowledge for the MCAT?

To answer some questions, you have uniform circular motion so you're going to have a constant acceleration inwards. You know your centripedal acceleration is v^2/r and its magnitude is constant and inwards. The only two other forces acting on your system are gravity and tension. Gravity is constant and downwards, so in order to maintain a constant centripedal acceleration the only thing that can change is the tension. Some numbers:

You have an acceleration of 30m/s^2 and it always points inwards. g = -10m/s^2.

acceleration due to tension = T

a = g + T

At the top of the circle you have gravity downwards in the same direction as centripetal acceleration. -30 = -10 + T. Your acceleration due to tension is 20m/s^2 downwards.

At the bottom you gravity still pointing downwards, and the acceleration is again inwards which in this case is upwards. Thus you have 30 = -10 + T. Your acceleration due to tension = 40m/s^2 upwards.

So yea, acceleration due to tension is highest at the bottom because it acts against gravity, and lowest at the top because it acts with gravity.

 

[TheBoondocks]

Thanks for the response. I'm still a bit confused though.

So, when we're talking about uniform circular motion (vertical):
1. Speed is constant
2. At the top, the tension on the rope is the lowest and at the bottom of the circle, the tension is the highest

When talking about non-uniform circular motion (vertical):
1. Speed isn't constant (highest at the bottom, lowest at the top). However, I don't see where the tangential acceleration is at the top. The Fgrav is pointing down so isn't there only one acceleration component (radial pointing towards the center)? If there is no tangential acceleration at the top, why is the speed the lowest at the top? Also, what is happening to the tension force at the top vs. at the bottom?

you are correct regarding uniform circular motion. lol, you understand it. it's the tangental acceleration that gives the increase in speed. at the top there is no tangental acceleration and this is true at the bottom as well. the speed is lowest at the top because the tangental acceleration is a function of mg. as it goes up, the tangental acceleration which is caused by gravity acts to slow it down, the tangental acceleration is opposite to the velocity. the reverse is true as it moves towards the bottom. Which is why it's SLOWEST at the top and fastest at the bottom. No difference in non uniform, the tension at the top is less since we have NO tangental. the top and bottom of NON-UNIFORM is the same as uniform. as far as components are concerned. WHAT changes is velocity the and the magnitude of tension the rest of the way.

 

[majik1213]

wrong spot sorry

 

[majik1213]

Is non-uniform circular motion required knowledge for the MCAT?

To answer some questions, you have uniform circular motion so you're going to have a constant acceleration inwards. You know your centripedal acceleration is v^2/r and its magnitude is constant and inwards. The only two other forces acting on your system are gravity and tension. Gravity is constant and downwards, so in order to maintain a constant centripedal acceleration the only thing that can change is the tension. Some numbers:

You have an acceleration of 30m/s^2 and it always points inwards. g = -10m/s^2.

acceleration due to tension = T

a = g + T

At the top of the circle you have gravity downwards in the same direction as centripetal acceleration. -30 = -10 + T. Your tension is 20m/s^2 downwards.

At the bottom you gravity still pointing downwards, and the acceleration is again inwards which in this case is upwards. Thus you have 30 = -10 + T. Your tension = 40m/s^2 upwards.

So yea, tension is highest at the bottom because it acts against gravity, and lowest at the top because it acts with gravity.

I think you mean "acceleration due to tension," T, not tension, which is mT, following your conventions.

 

[TheBoondocks]

In uniform circular motion, either there can be no gravity, or we must assume that the tension varies in a way that perfectly cancels out the gravitational influence of gravity on the net centripetal force. The answer to 1 is yes, and to 2 is yes, provided there is gravity.




correct; speed cannot remain constant, or else the motion becomes uniform by definition. At the top the tangential acceleration is pointing in the direction the object would travel if all the forces suddenly vanished. If you are, for example, watching the object rotate before you in a clockwise path, then at the top the tangential acceleration, if non-zero, points to the right.

In practice, you cannot make a rock on a rope go in uniform motion if you spin it vertically. To do so would require that you accelerate it tangentially at a max when it's halfway going up, and that you negatively accelerate it with equally maximal magnitude tangentially when it's halfway down. You must smoothly change this acceleration much as a cosine or sine wave. If you do that, then props.

that's wrong. at the top the tangental acceleration is zero. The velocity vector is to the right, but there is no tangental acceleration. At the top and bottom there is only radial acceleration. Everything else is correct. Uniform circular motion doesn't exist in the vertical direction. It can be done in the horizonatal but great care must be taken. however, if you mean't right after it moves from the top then that's correct.

 

[majik1213]

you are correct regarding uniform circular motion. lol, you understand it. it's the tangental acceleration that gives the increase in speed. at the top there is no tangental acceleration and this is true at the bottom as well.

I disagree, for a_t = r*a_c, where a_t is the tangential acceleration and a_c is the centripetal acceleration. (At the same time, I think your claim may still be correct, but for a different reason that I cannot say at the moment.)

the speed is lowest at the top because the tangental acceleration is a function of mg. as it goes up, the tangental acceleration which is caused by gravity acts to slow it down, the tangental acceleration is opposite to the velocity.

I think it's caused by the net centripetal force, of which the force of gravity is a component along with tension.

 

[majik1213]

that's wrong. at the top the tangental acceleration is zero. The velocity vector is to the right, but there is no tangental acceleration. At the top and bottom there is only radial acceleration. Everything else is correct. Uniform circular motion doesn't exist in the vertical direction. It can be done in the horizonatal but great care must be taken.

I see, could you please tell me though about that a_t = r * a_c thing though? I think there has to be some tangential acceleration to keep the rock at a constant speed, and this tangential acceleartion doesn't appear in uniform motion. By the way, by "great care" did you mean that when we spin a rock over our head in a horizontal plane, the motion becomes uniform in the limit that the acceleration due to gravity becomes insignificant compared to the magnitude of the tension pulling the rock down?

 

[TheBoondocks]

I disagree, for a_t = r*a_c, where a_t is the tangential acceleration and a_c is the centripetal acceleration. (At the same time, I think your claim may still be correct, but for a different reason that I cannot say at the moment.)

at=mgsin(theta) the acceleration tangential is a function of mg and the angle only.

I think it's caused by the net centripetal force, of which the force of gravity is a component along with tension.

here's a link, scroll down to pages 2 and 3. http://ocw.mit.edu/NR/rdonlyres/hs/physics/b/8_01_fall_2003_binder12.pdf

 

[TheBoondocks]

I see, could you please tell me though about that a_t = r * a_c thing though? I think there has to be some tangential acceleration to keep the rock at a constant speed, and this tangential acceleartion doesn't appear in uniform motion. By the way, by "great care" did you mean that when we spin a rock over our head in a horizontal plane, the motion becomes uniform in the limit that the acceleration due to gravity becomes insignificant compared to the magnitude of the tension pulling the rock down?

yes, you nailed it.

 

[TheBoondocks]

I disagree, for a_t = r*a_c, where a_t is the tangential acceleration and a_c is the centripetal acceleration. (At the same time, I think your claim may still be correct, but for a different reason that I cannot say at the moment.)



I think it's caused by the net centripetal force, of which the force of gravity is a component along with tension.

the equation is wrong. at=r *alpa alpa is the rate of change of angular speed. i think you got the rate of change of angular speed confused with the rate of change of angular velocity.

 

[RSAgator]

I think you mean "acceleration due to tension," T, not tension, which is mT, following your conventions.

fixed, thanks =)

 

[majik1213]

the equation is wrong. at=r *alpa alpa is the rate of change of angular speed. i think you got the rate of change of angular speed confused with the rate of change of angular velocity.

hmm interesting .. I usually designate the rate of change in angular speed simply as |alpha| .. what do you use to designate the rate of change in angular velocity (wrt time, of course)?

but I can see clearly that you're correct and I am wrong about the concept of acceleration tangentially.

 

[TheBoondocks]

hmm interesting .. I usually designate the rate of change in angular speed simply as |alpha| .. what do you use to designate the rate of change in angular velocity (wrt time, of course)?

but I can see clearly that you're correct and I am wrong about the concept of acceleration tangentially.

i'm in calc based physics. we call angular velocity omega.

 

[BloodySurgeon]

i'm in calc based physics. we call angular velocity omega.

good for you... want a cookie?

 

[TheBoondocks]

good for you... want a cookie?

i wasn't bragging. perhaps the reason for the different terms is that in algebra based physics they use different notations. that's why I mentioned i'm in calc based.

 

[BloodySurgeon]

i wasn't bragging. perhaps the reason for the different terms is that in algebra based physics they use different notations. that's why I mentioned i'm in calc based.

nah, i know.... if you knew me better u'd know I'm just messin with u