Register to reply

Physics Problem!

by physicsgirl101
Tags: physics
Share this thread:
physicsgirl101
#19
Jan9-05, 04:58 PM
P: 36
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..... ???
apchemstudent
#20
Jan9-05, 05:09 PM
P: 220
Quote 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...
physicsgirl101
#21
Jan9-05, 05:19 PM
P: 36
Thanks, definitely keep me posted, Thanks!!
vincentchan
#22
Jan9-05, 05:40 PM
P: 610
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...
physicsgirl101
#23
Jan9-05, 05:58 PM
P: 36
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..... ???
dextercioby
#24
Jan9-05, 06:00 PM
Sci Advisor
HW Helper
P: 11,915
The net acceleration is zero.It is composed however from 2 "pieces":one is due to gravity:[itex]\vec{g}[/itex] and the other is due to centripetal character of the tension force [tex] -\omega^{2} l\vec{r} [/tex],with
[tex] |\vec{r}| =1 [/tex].

If u see that
[tex] \omega=\sqrt{\frac{g}{l}} [/tex]
,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...
dextercioby
#25
Jan9-05, 06:03 PM
Sci Advisor
HW Helper
P: 11,915
Quote 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.
vincentchan
#26
Jan9-05, 06:07 PM
P: 610
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>?
vincentchan
#27
Jan9-05, 06:08 PM
P: 610
physicsgirl: did u see the formulas f=mv^2/r?
apchemstudent
#28
Jan9-05, 06:09 PM
P: 220
Quote 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...
dextercioby
#29
Jan9-05, 06:15 PM
Sci Advisor
HW Helper
P: 11,915
Quote 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...
vincentchan
#30
Jan9-05, 06:19 PM
P: 610
Quote Quote by dextercioby
centripetal character of the tension force [tex] -\omega^{2} l\vec{r} [/tex],with
dex:
[tex]\omega[/tex] is the angular speed, and it is not equal [tex]\sqrt{g/l}[/tex], indeed, it is not constant, At the highest point, [tex]\omega = 0 [/tex], on the other hand, [tex]\omega = maximun[/tex] at the lowest point. [tex]\sqrt{g/l}[/tex]is the average [tex]\omega[/tex] under a whole period, so, the tension on the string is greater than mg,,,,,
apchemstudent
#31
Jan9-05, 06:20 PM
P: 220
Quote 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....
physicsgirl101
#32
Jan9-05, 06:24 PM
P: 36
Quote 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
physicsgirl101
#33
Jan9-05, 06:25 PM
P: 36
Quote 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!
vincentchan
#34
Jan9-05, 06:26 PM
P: 610
mr dex, b4 your reply, think carefully, the faster the block move, the greater tension on the string......
apchemstudent
#35
Jan9-05, 06:27 PM
P: 220
Quote Quote by physicsgirl101
i've never heard of centrifugal OR centripetal forces!
the centripetal force is simply radial acceleration times the mass...
Pyrrhus
#36
Jan9-05, 06:32 PM
HW Helper
Pyrrhus's Avatar
P: 2,277
Actually, dextercioby is right, because when the bullet hits the block and it swings up to a theta max then the block will accelerate back to the lowest point because of gravity where it will only have max velocity, therefore the acceleration will be 0, and this is definetly a simple pendulum or mathematical pendulum where the dimensions of the block are not taken into account.


Register to reply

Related Discussions
Having a problem with a fairly early Physics with Calculus 1 problem Introductory Physics Homework 4
Help with this physics problem Introductory Physics Homework 2
Physics problem Introductory Physics Homework 1
Need help with some physics problem Introductory Physics Homework 2
Help...physics problem Introductory Physics Homework 4