Determine the speed and tension of keys swinging in a circular path

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Homework Statement

Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane.
a)Determine the slowest speed that the keys can swing and still maintain a circular path.
b)Determine the magnitude of the tension in the string at the bottom of the circle.

Relevant Equations

Fnet = Fg+ Ft
Fc = Fg + Ft
Fc= mv^2 / r
Fg = mg

a)Determine the slowest speed that the keys can swing and still maintain a circular path.

Fnet = Fg + Ft
Fc = Fg + Ft

When Ft = 0, Fc = Fg
So, Fc = mv^2 / r and Fg = mg

mv^2/r = mg

v = √gr
v = √9.81 * 0.25
v = 1.56 m/s

Therefore, the slowest speed that the keys can swing and still maintain a circular path is 1.56m/s.

b) Determine the magnitude of the tension in the string at the bottom of the circle.

In order to determine the magnitude of the tension in the string at the bottom of the circle I would need to find the minimum speed at the bottom of the circle. How can I find that? I know that I need to include velocity from part a, but not sure what else to do.

 

[Lnewqban]

The minimum velocity should happen at the top of the circle, and potential energy becomes increased velocity at the bottom of the circle, where you have two forces acting vertically and in the same direction.

 

[haruspex]

I know that I need to include velocity from part a, but not sure what else to do.

Can you think of a relevant conservation law?

 

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Can you think of a relevant conservation law?

I just re-read through all of my lessons and I couldn't find any conservation laws. Is there another way to solve this question? The law of conservation of energy maybe? I just haven't seen that get used in any examples in my textbook for questions like this.

 

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I saw someone online calculate the minimum speed at the bottom of the circle like this:
v=√v^2_top + 2gr
= √3 * v_top
= 2.71 m/s.
I don't really understand this equation, and I don't know what it's called or why it was used.

After this, to solve for the tension they did

Ft = mg + mv^2 / r

Ft = 3.92 N

I understand the tension part, just not how they solved for the minimum speed at the bottom of the circle.

 

[haruspex]

The law of conservation of energy maybe?

That's the one, and this is what @Lnewqban described in post #2.
It leads to this equation:

I saw someone online calculate the minimum speed at the bottom of the circle like this:
$v=\sqrt{v_{top}^2 + 2gr}$

 

[Traced]

That's the one, and this is what @Lnewqban described in post #2.
It leads to this equation:

Okay. Could you show me the steps I need to take to arrive at that equation? Or help me get started please!

Would I be using mv^2/r = mg?

 

[haruspex]

Would I be using mv^2/r = mg?

No, because at the bottom there will be tension.
As you have been told twice already, it's conservation of energy that you need.
At the top of the swing, the keys have kinetic energy and, by virtue of their height, gravitational potential energy. As they descend, they lose GPE but get faster. Other than a tiny bit of air resistance, their total energy stays constant.
At the top of the arc, how much KE do they have?
In descending from the top of the arc to the bottom, how much GPE do they lose?
So what is their KE at the bottom?

 

[hmmm27]

The phrase "minimum speed at the bottom of the circle" is a bit confusing. Technically, that would be zero. So, how about "speed at the bottom of the circle, assuming minimum speed at the top". Yeah, I know, same thing.

From there... at the top you've PE (vertical) and KE (horizontal) ; at the bottom, no PE right ? just KE(/RE).

So, where did the PE go ?