Vertical Circular Motion Question? Halp!

[PB]

Q :   A yo-yo is swung with a constant speed in a vertical
circle.
At which point in the circular path is there the
lowest amount of tension in the string? Why?

*I get that the lowest tension occurs when the yoyo is at the top, but how do I explain it? the textbook answer makes no sense to me (At the top of its path, the yo-yo has an downwards acceleration and so the net force is down. This indicates that the tension force is less than Fg.) 

Thanks! (+1)

[Homer]

im not too sure but, when the yoyo is at the bottom the gravity pulls it down and the centripetal force pulls it up. now when you have something that is given a force in opposite sides, like stretching a string i guess, the tension is greater.

when it is up, the centripetal force aswell as the gravityy is down; both in the same direction hence I THINK the tension would be less!

[PB]

mmmm, i dunno about that...  How I thought about it is derived from the formula Fnet=Ftension + Fg.
Seeing as the yoyo is travelling at a constant speed, that would theoretically mean that its centripetal acceleration would also be constant throughout (based on the formula a=v^2/r). Hence, the Fnet would also be constant throughout the circular path (based on the formula F=ma).
Now we can ascertain two main things:
1. we have established that Fnet and Fg is constant throughout, that means that the only variable is Ftension.
2. Given that the yoyo is doesn't fall down when at the top of swing, we can deduce that Fnet is greater than or equal to Fg

So I gave Fnet an arbituary no. (Say 3)  and Fg (1).

At the bottom of the swing.
Fnet would be 3 up, Fg would be 1 down.   
so using the formula  Fnet=Ftension + Fg and subbing in values. (Assuming the up is positive direction)
3=Ftension + (-1)                  <------(up is + direction so Fg would be -1)
Ft=4

so Ftension has a magitude of 4 going up.

At the top of the swing.
-3= Ft -1
Ft= -2
So Ftension has a magnitude of 2 going down. 

Hence, Ftension is at its lowest when the yoyo is at the top.

Yeh? What do you think of my understanding?

[Robert123]

I would answer it with these dot points (note: vertical motion isn't on the study design this year)
To maintain its speed, the net force must be constant
When the ball is at the top, gravity compliments tension resulting in less tension.
When the ball is at the bottom, gravity opposes tension resulting in more tension to maintain the centrepetal force.
Therefore, the string is going to break at the bottom due to a higher tension

[PB]

(note: vertical motion isn't on the study design this year)

Thanks for your answer.

Although I beg to differ about it not being included in the study deisng. this is a quote from VCAA Physics 2013-2016 Study Design "apply Newton’s second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only"

So i think its still in the study design but just not as detailed?

[lzxnl]

The radially inward centripetal acceleration v^2/r is a requirement for an object to move in a circle with that speed and radius. It does not matter if the object is in uniform circular motion or not. If the object's angular velocity is increasing, that just means that there is a force component acting parallel to the object's motion as well. The centripetal acceleration, however, is still v^2/r.
If we take a snapshot of the object's motion at the top of the circle, and taking down to be positive, this is what we have:
F net=T+mg as both the weight force and the tension force are directed straight down.
This means T=F net-mg=mv^2/r-mg

At the bottom of the circle, T is now directed upwards. So is the centripetal force.
Taking up to be positive, we have F net=T-mg, T=F net +mg=mv^2/r+mg

I think the answer to your question should be clearer now.

[PB]

Yep, I think I understand now. Thanks nliu1995!