Question

A 0.050 Kg yo-yo is swung in a vertical circle on the end of its 0.30 m long string at the slowest speed that the yo-yo can have.

A) What is the minimum speed of the yo-yo at the bottom of the circular path to keep the circular motion.

B) What’s the minimum velocity of the yo-yo at the top of the yoyo’s circular to maintain its circular path?

Answer 1

I’m here

Answer 2

A quick glance at my notes shows the use of:

\[\Sigma F = ma _{c} = T + mg\]

\[a _{c} = \frac{ V^2 }{ r }\]

So you can find velocity

Answer 3

But i don’t have the velocity?

Answer 4

What happened?

Answer 5

Yes

Answer 6

Wait.... if the yo-yo is at the bottom...wouldn’t that make the bottom side of the circle final velocity?

Answer 7

We are solving for the minimum velocity at the bottom of the circle fir.st

Answer 8

Energy is relative. As are most things in classical physics.


we look at total mechanical energy (E) at the bottom of the cycle:

\(E = \frac{1}{2} m v_i^2\)

Answer 9

Ok sir

Answer 10

To be clearer:


\(E_1 = \frac{1}{2} m v_1^2\)

Answer 11

Initial energy?

Answer 12

Now, at the top we have total energy \(\frac{1}{2}m v_2^2 + \text{ "some potential energy"}\)

Answer 13

At the top of the circular path, we have gravitational potential energy. No?

Answer 14

we have increased the gravitational potential energy of the yo-yo

Answer 15

Ok

Answer 16

by how much have we increased it's GPE?

Answer 17

Twice since there’s 2r? Idk :/

Answer 18

🤓

GPE is \(mgh\)

so from the bottom of the circle to the top the particle gains this potential energy: \(m g (2r)\)

I think that is what you meant

Answer 19

Yeah. Lol

Answer 20

you are lost, right ?!

Answer 21

Kinda...

Answer 22

Oh so that’s where the 2r came from... i was having hard time trying to figure out where did that 2r come from. So GPE = meh got it .

Answer 23

OK, so I am guessing you've been given some formulae for acceleration in a circle.

like \(a = - \dfrac{v^2}{r}\)

so at the top of the circle you say that :

\(a = - \dfrac{v^2}{r} = g = 0\)

Answer 24

Mechanical energy is made up of two forces. potential energy which is the gravitational potentional energy and the kinetic energy which is the 1/2 mv^2/r

Answer 25

Yeah that’s the formula i’ve Been given but go on.

Answer 26

typo:


\(a = - \dfrac{v^2}{r} - g = 0\)

Answer 27

ignore the minus signs. think about forces and inertia [ie the reluctance of mass to comply with forces]

Answer 28

Ok so \[A = v^2/r + g ?\]

Answer 29

If we take the negative signs out?

Answer 30

I’m here daddy :)

Answer 31

For a point mass moving along a linear 1-D gravitational field, we can certainly say that:

\(\dfrac{d}{dt} \left( \frac{1}{2} m v^2 + m gh\right) = 0\)

cos we can say that energy is conserved

Answer 32

none of this answers yer question though!

Answer 33

Hey i say yer! :D

Answer 34

Yeah so how do we proceed from here?

Answer 35

So do we just plugin the numbers and solve for the velocity?

Answer 36

not using your equation. use some proper equations. let me show you !!

Answer 37

So We haven’t even made a proper equation yet?

Answer 38

in your mind, i think we have. trust me.

Answer 39

Ok so where do we go from here?

Answer 40

Jesus what am i gunna do if this question shows up in the exams? Uhg :/

Answer 41

Well, (b) is zero, ...... plucky troller, ...., and you can get (a) from the other stuff

yah!

Answer 42

Other stuff? Jesus

Answer 43

Oh i get it

Answer 44

You found Jesus ?!?!

Answer 45

Like, in there?!?!

Answer 46

Wait what do you mean by....”(b) is zero” and no i didn’t find jesus. Jesus i’m Not even Christian

Answer 47

just teasing

chillax and stop trolling

Answer 48

Ok

Answer 49

i think you should come out, too

Answer 50

Ok I’m here

Answer 51

awesome

Answer 52

So again.... we have 1/2 mv^2 + meh = 0

Answer 53

Where do we go from here?

Answer 54

are you seriously challenging me in a troll-war ?!?!

you are Putin. OK?

Answer 55

What? I just want my help. That’s it.

Answer 56

Sorry i made a mistake up there....mgh*

Answer 57

Are you there?

Answer 58

You got trolled?

Answer 59

What? I thought i was getting help

Answer 60

And I think he was legit help.