Circulation Motion BR Physics

[Jay2910]

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Hello everyone,

I have a question about Passage II on the Practice section( Its page 101 and in the "practice passages" section).

15) The question asks if a ball in experiment 2 is swung in a uniform circular motion then how are the magnitudes of tensions at points A,B,C and D in figure 1 related?

The answer choices are:
a)Ta=Tc, Tb<Td
b) Ta=Tb=Tc=Td
c) Ta<Tb<Tc<Td
d) Ta<Tc,Tb=Td

According to the explanation, the answer is D. Here are my questions
1) There is absolutely no tension on the string when its on point A( top of the circle) right? Cause aren't you using your momentum to get it up there in the first place?

2) I thought at point C, Tnesion and mg must be direct opposites and thus cancel each other out . .why is that notion wrong?So why do they have Tc-mg=(mv^2)/r? Why isn't point C like point B or point D?

3) At points B and D . .why is the tension directly equal to the (mv^2)/r? Is it because tension and mg are perpendicular to each other?

4) Why isn't tension the same all throughout the circle? I thought, that the only outside force it was encountering was gravity

 

[Jay2910]

People please! Some help would be great cause I'm quite desperate! Am I really missing something on this question? Is there something that you all saw that I did not?

 

[MedChad]

Hello everyone,

I have a question about Passage II on the Practice section( Its page 101 and in the "practice passages" section).

15) The question asks if a ball in experiment 2 is swung in a uniform circular motion then how are the magnitudes of tensions at points A,B,C and D in figure 1 related?

The answer choices are:
a)Ta=Tc, Tb<Td
b) Ta=Tb=Tc=Td
c) Ta<Tb<Tc<Td
d) Ta<Tc,Tb=Td

According to the explanation, the answer is D. Here are my questions
1) There is absolutely no tension on the string when its on point A( top of the circle) right? Cause aren't you using your momentum to get it up there in the first place?

2) I thought at point C, Tnesion and mg must be direct opposites and thus cancel each other out . .why is that notion wrong?So why do they have Tc-mg=(mv^2)/r? Why isn't point C like point B or point D?

3) At points B and D . .why is the tension directly equal to the (mv^2)/r? Is it because tension and mg are perpendicular to each other?

4) Why isn't tension the same all throughout the circle? I thought, that the only outside force it was encountering was gravity

1. Calm down
2. Lets try again

1. At point A, I'm not sure if its zero, but you can conclude that it is less then point c. Think about it if you had a yoyo or something and started swinging it in a circle. At the top, their is a little give in the rope, therefore your tension is small. At the bottom (point c) you have all of the forces going down with it, so the tension is going to be the greatest here.
2. Point c is not like points b or d since all of the forces are parallel and together, where at points b and d you still have some of your force pointing inwards to the center. So their tension is greater than a, but less than point c.
3. I think your right but I would like someone else to try and verify that for you.
4.Tension isn't the same all around because gravity only goes down right? So therefore if you were swinging the ball in the downward arc, gravity is going along with you, but if you are swinging it upward, then gravity is against you, therefore changing your tension in the string. Hopefully some of this helped a little

 

[tdog60tf]

Hello everyone,

I have a question about Passage II on the Practice section( Its page 101 and in the "practice passages" section).

15) The question asks if a ball in experiment 2 is swung in a uniform circular motion then how are the magnitudes of tensions at points A,B,C and D in figure 1 related?

The answer choices are:
a)Ta=Tc, Tb<Td
b) Ta=Tb=Tc=Td
c) Ta<Tb<Tc<Td
d) Ta<Tc,Tb=Td

According to the explanation, the answer is D. Here are my questions
1) There is absolutely no tension on the string when its on point A( top of the circle) right? Cause aren't you using your momentum to get it up there in the first place?

2) I thought at point C, Tnesion and mg must be direct opposites and thus cancel each other out . .why is that notion wrong?So why do they have Tc-mg=(mv^2)/r? Why isn't point C like point B or point D?

3) At points B and D . .why is the tension directly equal to the (mv^2)/r? Is it because tension and mg are perpendicular to each other?

4) Why isn't tension the same all throughout the circle? I thought, that the only outside force it was encountering was gravity

1. There is some amount of tension in the string. The ball tied to the string is whats causing the tension to exist. Ta= m (v^2/r)

2. The forces do not cancel each other out. Look at the equation. Fc = Tc-mg=m(v^2/r) or Tc= M (v^2/r + mg)

3. Correct do a force diagram. T-0=m(v^2/r) & mg -0 = m(V^2/r)
t= m(V^2/r) mg = M(V^2/r)
The forces are the same for the sides.....and are equal....therefore erroneous to the question being asked.


4. it is different due to centripetal acceleration. the top has Ta= m(V^2/r) -mg while bottom has tc= m(V^2/r) +mg. Therefore more tension at the bottom.



Do force diagrams and work these out i think that will help.

 

[dranet]