Understanding Simple Gravity Pendulum - Urgent Help Needed

AI Thread Summary
The discussion focuses on understanding the dynamics of a simple gravity pendulum, particularly the derivation of formulas related to its motion. Key points include the relationship between the forces acting on the pendulum, specifically the tension in the string and gravitational force, and how these forces change with the angle of displacement. The participant questions why the net force and angle relationships differ when the pendulum is displaced, and they seek clarification on the conditions under which the motion becomes harmonic. Additionally, there is confusion regarding the use of sine approximations for small angles and the implications for acceleration direction. The importance of free body diagrams is emphasized as a tool for visualizing and solving these problems.
kajalove
Messages
15
Reaction score
0
hi

I know about harmonic oscilation, but I'm having trouble understanding how we derived formulas for gravity pendulum. Please read on.



If a ball on a string ( string is attached to the ceiling ) is displaced from its equilibrium position by angle A1, then forces on this ball are force of string F[v] and F[g].

F[v1] ... component of F[v] parallel to F[g] and of opposite direction to F[g]

http://img473.imageshack.us/img473/1854/nihaloje2.th.png

BTW - if picture doesn't show up then please look at the attached jpg file





1)

a)
Now why ( when angle A1 > 0 ) isn't the magnitude of F[v1] equal to F[g] --> F[v1] = -F[g]?


b)
I assume bigger the angle A, smaller is F[v1]. Why?
I assume it's because F[v] is constant no matter what the angle A is, but why is that?





2)
According to my book angles A and A1 are the same:

Code:
[B]F[net] = m * g * sin[A1]  =  m * g * sin[A][/B].

I'd imagine angle A being the same as angle A1 only if F[g] = F[v1]. Then direction of F[net] would be horizontal. But since that is not the case thus the two angles shouldn't be the same.





3)
I will quote my book:

Distance of a ball from equilibrium state can be stated with

Code:
[B]L = A1 * d = A * d [/B]
, where L is arc of a circle. When at angle A, the net force on the ball is F = m * g * sin[A], which gives the ball acceleration
Code:
[B]a = - g * sin[A][/B]

Acceleration vector a certainly isn't linear with L = A * d, and thus the osciliation isn't harmonic. But it becomes harmonic, if angle A is small enough for us to replace sin[A] with A




a)
a certainly isn't linear with L = A * d
I assume by that they mean to say that when arc L is twice as great, a isn't twice as great.
But what has that got to do with harmonic osiclation? Is with harmonic oscillation a linear with L?
Can you show me some proof of that?





b)
But it becomes harmonic, if angle A1 is small enough for us to replace sin[A1] with A1
First of all, I'm not sure that sin[A1] and A1 are ever roughly the same size, since no matter how small A1 is, sin[A1] will always be 100 or more times smaller. Right?




c)
Second, even if sin[A1] and A1 have about the same value when A1 is small enough, what is the purpose of replacing sin[A1] with A1? Why do we want to do that?





d)
Also, why is acceleration vector a negative?
I realize that when a has opposite direction to ball's velocity that it has to be negative. But sometimes ball's velocity and acceleration vectors have same direction and thus a should be positive?



cheers
 

Attachments

  • nihalo.jpg
    nihalo.jpg
    7.1 KB · Views: 473
Last edited by a moderator:
Physics news on Phys.org
Most of your questions can be answered by drawing a free body diagram. For question (a) consider what would happen if Fv1 = -Fg?
 
Hootenanny said:
Most of your questions can be answered by drawing a free body diagram. For question (a) consider what would happen if Fv1 = -Fg?

Ball would start moving in a horizontal direction. I realize that and I know this is not the case, but why doesn't it happen?!
 
Um if your angle keeps increasing it eventually reaches 90, so it keeps getting smaller and when it reaches 90 it goes to zero and tension is only determined in x direction since your tension component is basically Tension*cos(theta). I hope that answered the angle part of your problem
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top