How Do You Solve Complex Pendulum Equations in Physics Homework?

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To solve complex pendulum equations in physics homework, it's essential to apply the principles of mechanical energy conservation to find the speed at various points, such as point D. The height at point D must be determined relative to a chosen reference point, which affects the potential energy calculations. The tension in the string can be calculated by considering the centripetal force and gravitational force acting on the mass at point D. It's crucial to clarify the direction of forces and ensure that kinetic energy is treated as a non-negative value, regardless of the motion's direction. Understanding the relationship between forces and the geometry of the pendulum is key to solving these problems effectively.
Helly123
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Homework Statement



For number 3,4,5
IMG_1092.png


IMG_1093.png


https://s4.postimg.org/qbp3xzq65/IMG_1092.png

https://s22.postimg.org/u220j60sx/IMG_1093.png

Homework Equations


Centripetal force = m v^2/R
T = centripetal force + w
Sigma F = m.a

The Attempt at a Solution


Number 3. Find the v at D point using mechanic energy eternity
While kinetic energy at A point is zero, and the potential energy is max (height max)
So, 1/2mvD^2 + mg.hD = 1/2mvA^2 + mg.hA
i don't know the height at D point...

After i find the v at D
I can find T
Which T = centripetal Force + weight
T = mvD^2/R + m.g

Number 4 almost the same
The string doesn't bend (what exactly it means...?)
So, centripetal force > weight??
m.vD^2/R > m.g ...(1)
But why the centripetal at point D has opposite direction than weight force? Is it because the string stuck at the nail at C point??

Find the v at D using mechanical energy eternity then add v to ...(1)

Number 5 i still don't know...
 

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Hello Helly, :welcome:

Unfortunately your pictures don't show up in your post. I get:

upload_2017-6-8_13-33-27.png


Can you try again ?
 
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BvU said:
Hello Helly, :welcome:

Unfortunately your pictures don't show up in your post. I get:

View attachment 205082

Can you try again ?
ok.. thanks for report. I've changed the pic.. :)
 
Helly123 said:
So, 1/2mvD^2 + mg.hD = 1/2mvA^2 + mg.hA
with ##v_A = 0 ##.
i don't know the height at D point...
make a sketch of the situation and you know it. Note that CB = CD.
 
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Helly123 said:
i don't know the height at D point
You know the length of the string, and you know how far C is below A. How far above C is D?
 
is the height of D start from C or B ...?

CD = a-b
BC = a-b
if D start from C then height of D = a-b
if D start from B then height of D = 2(a-b)
 
Check your energy conservation equation: If A is at height zero, you must take height of D wrt the same referernce.
 
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Helly123 said:
is the height of D start from C or B ...?
BvU said:
Check your energy conservation equation: If A is at height zero, you must take height of D wrt the same referernce.
if A = height zero.. then kinetic energy is max...
then B point the kinetic energy is zero... how can it be...
 
It was you who chose height A = 0 ! Therefore at point B the height is -a and the energy conservation still reads $$
{1\over 2} mv_B^2 + m g h_B = {1\over 2}mv_A^2 + mg\,h_A \quad {\rm or} \\
{1\over 2} mv_B^2 - mga = 0 $$
 
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  • #10
While kinetic energy at A point is zero, and the potential energy is max (height max)
BvU said:
It was you who chose height A = 0 ! Therefore at point B the height is -a and the energy conservation still reads $$
{1\over 2} mv_B^2 + m g h_B = {1\over 2}mv_A^2 + mg\,h_A \quad {\rm or} \\
{1\over 2} mv_B^2 - mga = 0 $$
BvU said:
It was you who chose height A = 0 ! Therefore at point B the height is -a and the energy conservation still reads $$
{1\over 2} mv_B^2 + m g h_B = {1\over 2}mv_A^2 + mg\,h_A \quad {\rm or} \\
{1\over 2} mv_B^2 - mga = 0 $$
hmm I see..
then it said the string doesn't bend, how that means Sir??
 
  • #11
As long as you have tension in a string, it does not bend but remains taut.
First work out number 3, once you have that, then number 4 is easy

By the way, what did you get for number 2 ?

Do you realize image 1093 is extremely fuzzy ?
 
  • #12
BvU said:
As long as you have tension in a string, it does not bend but remains taut.
First work out number 3, once you have that, then number 4 is easy

By the way, what did you get for number 2 ?

number 2? I get sqrt 2ga. do you even know the questions Sir.. but I'm still trying for number 3...
 
  • #13
Very unsharp, in image 1093 under (2) It says: Find the magnitude of the tension in the string just before the small ball reaches B.
##\sqrt{2ga}\ ## is not among the choices. But also, 2mg ...
Is the wrong answer (as I already suspected :rolleyes: ) .

You need to get this right: the same issue also needed in (3).
 
  • #14
BvU said:
Very unsharp, in image 1093 under (2) It says: Find the magnitude of the tension in the string just before the small ball reaches B.
##\sqrt{2ga}\ ## is not among the choices. But also, 2mg ...
Is the wrong answer (as I already suspected :rolleyes: ) .

You need to get this right: the same issue also needed in (3).
lol.. sorry Sir. I meant 3mg..
 
  • #15
Agree with 3mg.
 
  • #16
BvU said:
Agree with 3mg.
oh... it's my mistake... lol i'll try again number 3.. how do you even know the question for number 2, from the pic there's no number 2
 
  • #19
I know that.
 
  • #20
BvU said:
I know that.
I don't even understand... How to solve number 3...
 
  • #21
How to solve number 3...?
 
  • #22
At point D you know the speed from energy conservation.
So you know ##mv^2/r##.
Take gravity into account and then you have T.
 
  • #23
BvU said:
At point D you know the speed from energy conservation.
So you know ##mv^2/r##.
Take gravity into account and then you have T.
No the problem is idk the speed.. And not sure the height of D. If the string not bend, it's possible if only the string stuck at nail on C point then goes around to D, it means the height of D is a-b. The v^2 at D = 2gb.. Then.. I don't find the answer match the options
 
  • #24
Helly123 said:
it means the height of D is a-b
with respect to what reference point ?
Helly123 said:
The v^2 at D = 2gb
How do you calculate that ?
 
  • #25
BvU said:
with respect to what reference point ?
How do you calculate that ?
Do you know the answer already?
 
  • #26
Helly123 said:
Do you know the answer already?
Helly123 said:
Do you know the answer already?
CB = a-b
CD =CB
CD = a-b.
 
  • #27
Suppose a is 0.5 m and b is 0.4 m. What is the height of D ?
What is the formula for the height of D with your reference point O = 0 ?
 
  • #28
BvU said:
Suppose a is 0.5 m and b is 0.4 m. What is the height of D ?
What is the formula for the height of D with your reference point O = 0 ?
Since the ball goes around to D point, the reference i used no longer O=0 but C.. Cuz the T is at C.. No i used is at O the height is a (max)
 
  • #29
Helly123 said:
Since the ball goes around to D point, the reference i used no longer O=0 but C.. Cuz the T is at C.. No i used is at O the height is a (max)
we know that OC = b (0.4 m)
Then CB = a-b (0.5-0.4)m
Then CB = 0.1m
CD = 0.1m. Then use CD as height of D
 
  • #30
Can someone give me hint for number 5?
 

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