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#19
Jan905, 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..... ??? 


#20
Jan905, 05:09 PM

P: 220




#21
Jan905, 05:19 PM

P: 36

Thanks, definitely keep me posted, Thanks!!



#22
Jan905, 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...



#23
Jan905, 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..... ???



#24
Jan905, 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... 


#25
Jan905, 06:03 PM

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P: 11,915

Daniel. 


#26
Jan905, 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>? 


#27
Jan905, 06:08 PM

P: 610

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



#28
Jan905, 06:09 PM

P: 220




#29
Jan905, 06:15 PM

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P: 11,915

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... 


#30
Jan905, 06:19 PM

P: 610

[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,,,,, 


#31
Jan905, 06:20 PM

P: 220




#32
Jan905, 06:24 PM

P: 36

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


#33
Jan905, 06:25 PM

P: 36




#34
Jan905, 06:26 PM

P: 610

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



#35
Jan905, 06:27 PM

P: 220




#36
Jan905, 06:32 PM

HW Helper
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.



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