## Physics Problem!

 Quote by dextercioby Hopefully i'm not mistaking...Weren't u that guy that started the thread:"Redemption:I challange Dextercioby"...??? If so,and you still don't believe that my result is correct,and apparently noone interviens in our little "polemics",i advise to start another thread with the title :"Redemption:I challange Dextercioby,AGAIN!!!!!!!!!!" and get some other opinions on it. In case u do that,here's my argument:"The tension force for a circular/mathematical pedulum is a centripetal force,and it the lowest point of the trajectory,the tension is exactly balancing the gravity force and therefore the body is in dynamical equilibrium,meaning the total force acting on it is zero.Since the acceleration is zero,the velocity has a critical point,which in this case is a maximum.So the maximum tension force is equal tot the total gravity,i.e. $$|\vec{T}_{max}|=(M+m)g$$ ." I rest my case... Daniel.
i hope some one else will check this out.... I've done a problem like this before concerning the ballistic pendulum... And why won't you agree with my calculation? Everything is in the ideal sense....

Gravitational energy is conserved so you can determine the velocity at the bottom of the trajectory.... which you can use to determine the Fc and Ft....
It's obvious that Fc is not 0 as i've said...

 First..... there is obviously major disagreement betweek dextercioby and apchemstudent, so does a third party want to give an opinion? Second.... My professor has never even mentioned the word "centrifugal force"----so unless there's another name for it that my professor has used and i'm just not making the connection, I doubt he would give us a problem involving it..... ???

 Quote by physicsgirl101 First..... there is obviously major disagreement betweek dextercioby and apchemstudent, so does a third party want to give an opinion? Second.... My professor has never even mentioned the word "centrifugal force"----so unless there's another name for it that my professor has used and i'm just not making the connection, I doubt he would give us a problem involving it..... ???
Yes i agree there should be a third party to check this out... I've already sent this problem out to a different website... I'll keep you posted on what happens...

 Thanks, definitely keep me posted, Thanks!!
 Hello, I am the third party you guys looking for . I am a big fans of dextercioby. he is very very good at physics. but, sadly, I must say he is WRONG this time, If you review the question carefully, you will see at the lowest point of the trajectory, accelaration is not zero.... Its speed doesn't change, but ITS VELOCITY (direction) does..... I will vote for apchemstudent this time...
 Again, i've never heard the word "centripetal force"----so unless there's another name for it that my professor has used and i'm just not making the connection, I doubt he would give us a problem involving it..... ???
 Blog Entries: 9 Recognitions: Homework Help Science Advisor The net acceleration is zero.It is composed however from 2 "pieces":one is due to gravity:$\vec{g}$ and the other is due to centripetal character of the tension force $$-\omega^{2} l\vec{r}$$,with $$|\vec{r}| =1$$. If u see that $$\omega=\sqrt{\frac{g}{l}}$$ ,u'll understand why in the lowest point of the trajectory the accleretion is zero and the velocty is maximum. Daniel. PS.For the "fan" part...

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 Quote by physicsgirl101 Again, i've never heard the word "centripetal force"----so unless there's another name for it that my professor has used and i'm just not making the connection, I doubt he would give us a problem involving it..... ???
Voilą,it proves my point.U don't need to know that the tension has a centripetal character.The second law of Newton would do it...

Daniel.

 the gravity can't balance the tension force If the accelaration is zero, why does the block changes its velocity (direction) after hitting the lowest point>?
 physicsgirl: did u see the formulas f=mv^2/r?

 Quote by vincentchan the gravity can't balance the tension force If the accelaration is zero, why does the block changes its velocity (direction) after hitting the lowest point>?
umm... its speed does change due to conservation of energy and the Fc is not constant... but i agree the direction also changes... This is what i'm trying to explain dextercioby... hope you have better luck...

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 Quote by vincentchan physicsgirl: did u see the formulas f=mv^2/r?

I'm sure she didn't.She said not to have been taught about circular motion and centripetal and centrifugal forces.

Daniel.

PS.Instead of asking questions,why don't u show us how u would apply the second principle of Mr.Newton...

 Quote by dextercioby centripetal character of the tension force $$-\omega^{2} l\vec{r}$$,with
dex:
$$\omega$$ is the angular speed, and it is not equal $$\sqrt{g/l}$$, indeed, it is not constant, At the highest point, $$\omega = 0$$, on the other hand, $$\omega = maximun$$ at the lowest point. $$\sqrt{g/l}$$is the average $$\omega$$ under a whole period, so, the tension on the string is greater than mg,,,,,

 Quote by dextercioby I'm sure she didn't.She said not to have been taught about circular motion and centripetal and centrifugal forces. Daniel. PS.Instead of asking questions,why don't u show us how u would apply the second principle of Mr.Newton...
She's just not used to centrifugal forces.... Same thing with me, my teacher told me there's no such thing except that people get mixed up with the centrifugal force for centripetal....

 Quote by vincentchan physicsgirl: did u see the formulas f=mv^2/r?
i've seen radial acceleration = v^2/r.... so yes, i guess, because that's just subbing a = v^2/r into f=ma

right? so yes, we've done some circular motion stuff.... but just using that equation

 Quote by apchemstudent She's just not used to centrifugal forces.... Same thing with me, my teacher told me there's no such thing except that people get mixed up with the centrifugal force for centripetal....
i've never heard of centrifugal OR centripetal forces!

 mr dex, b4 your reply, think carefully, the faster the block move, the greater tension on the string......