How Do You Solve Complex Pendulum Equations in Physics Homework?

[Helly123]

Homework Statement

For number 3,4,5

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...

 

[BvU]

Hello Helly,

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

Can you try again ?

 

[Helly123]

Hello Helly,

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.. :)

 

[BvU]

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.

 

[haruspex]

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?

 

[Helly123]

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)

 

[BvU]

Check your energy conservation equation: If A is at height zero, you must take height of D wrt the same referernce.

 

[Helly123]

is the height of D start from C or B ...?

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...

 

[BvU]

It was you who chose height A = 0 ! Therefore at point B the height is -a and the energy conservation still reads $$
{1\over 2} mv_B^2 + m g h_B = {1\over 2}mv_A^2 + mg\,h_A \quad {\rm or} \\
{1\over 2} mv_B^2 - mga = 0 $$

 

[Helly123]

While kinetic energy at A point is zero, and the potential energy is max (height max)

It was you who chose height A = 0 ! Therefore at point B the height is -a and the energy conservation still reads $$
{1\over 2} mv_B^2 + m g h_B = {1\over 2}mv_A^2 + mg\,h_A \quad {\rm or} \\
{1\over 2} mv_B^2 - mga = 0 $$

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??

 

[BvU]

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 ?

 

[Helly123]

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...

 

[BvU]

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 ) .

You need to get this right: the same issue also needed in (3).

 

[Helly123]

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 ) .

You need to get this right: the same issue also needed in (3).

lol.. sorry Sir. I meant 3mg..

 

[BvU]

Agree with 3mg.

 

[Helly123]

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

 

[BvU]

how do you even know the question for number 2, from the pic there's no number 2

From the attached image https://www.physicsforums.com/attachments/img_1093_1-png.205087/

But, as I said, it's very fuzzy !

 

[Helly123]

From the attached image https://www.physicsforums.com/attachments/img_1093_1-png.205087/

But, as I said, it's very fuzzy !

I already edit the picture on the thread.. https://s22.postimg.org/u220j60sx/IMG_1093.png

 

[BvU]

I know that.

 

[Helly123]

I know that.

I don't even understand... How to solve number 3...

 

[Helly123]

How to solve number 3...?

 

[BvU]

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.

 

[Helly123]

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

 

[BvU]

it means the height of D is a-b

with respect to what reference point ?

The v^2 at D = 2gb

How do you calculate that ?

 

[Helly123]

with respect to what reference point ?
How do you calculate that ?

Do you know the answer already?

 

[Helly123]

Do you know the answer already?

Do you know the answer already?

CB = a-b
CD =CB
CD = a-b.

 

[BvU]

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 ?

 

[Helly123]

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)

 

[Helly123]

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

 

[Helly123]

Can someone give me hint for number 5?

 

[BvU]

Number 3 first. Write down clearly your energy conservation equation.

 

[Helly123]

Number 3 first. Write down clearly your energy conservation equation.

This how i did it https://s4.postimg.org/ta19uxcz1/image.jpg

 

[BvU]

Unconventional presentation, but (quite !) good. So, on to number (5):

$mv^2\over r$ of tension is needed to maintain the circular motion.
Once the component of $mg$ along the wire exceeds $mv^2\over r$ the circular motion is interrupted. This happens somewhere between B and D, at a height above C.
Introduce some coordinate, e.g. $\theta$ and work out $v(\theta)$ and from that $T(\theta)$.

 

[Helly123]

Unconventional presentation, but (quite !) good. So, on to number (5):

$mv^2\over r$ of tension is needed to maintain the circular motion.
Once the component of $mg$ along the wire exceeds $mv^2\over r$ the circular motion is interrupted. This happens somewhere between B and D, at a height above C.
Introduce some coordinate, e.g. $\theta$ and work out $v(\theta)$ and from that $T(\theta)$.

thanks Sir. that's the most effective way for me to fulfill your order. btw, on number 3-3 I set KE for D negative, because it goes to the left. is that ok??

 

[BvU]

That's what I get for looking at the answers too summarily. Sorry.

No. Kinetic energy is never negative (*). The potential energy at D is not $mg(2b-a)$ but $mg(a-2b)$ because the height at D is lower than the height at A.

(*)
Both $m$ and $v^2$ are non-negative. For kinetic energy there is no direction, so 'because it goes to the left' does not apply.

 

[Helly123]

That's what I get for looking at the answers too summarily. Sorry.

No. Kinetic energy is never negative (*). The potential energy at D is not $mg(2b-a)$ but $mg(a-2b)$ because the height at D is lower than the height at A.

(*)
Both $m$ and $v^2$ are non-negative. For kinetic energy there is no direction, so 'because it goes to the left' does not apply.

ok somehow maybe I understand. i'll try again then

 

[Helly123]

That's what I get for looking at the answers too summarily. Sorry.

No. Kinetic energy is never negative (*). The potential energy at D is not $mg(2b-a)$ but $mg(a-2b)$ because the height at D is lower than the height at A.

(*)
Both $m$ and $v^2$ are non-negative. For kinetic energy there is no direction, so 'because it goes to the left' does not apply.

ok somehow maybe I understand. i'll try again then

 

[Helly123]

How do you know its above height C when mg exceeds mv^2/R ? And can you give me another hint to work on tetha to find v?

 

[haruspex]

How do you know its above height C when mg exceeds mv^2/R ? And can you give me another hint to work on tetha to find v?

Draw a free body diagram with the mass somewhere between heights C and D, string still taut. Let the string make angle theta to the vertical here.
What is the speed of the mass, as a function of theta?
Write the ΣF=ma equation for the vertical direction.

 

[Helly123]

Draw a free body diagram with the mass somewhere between heights C and D, string still taut. Let the string make angle theta to the vertical here.
What is the speed of the mass, as a function of theta?
Write the ΣF=ma equation for the vertical direction.

How if i used ΣF = ma on horizontal? Or i make all force in tetha direction?

 

[haruspex]

Or i make all force in tetha direction?

Yes, sorry, that's better. ΣF=ma along the direction of the string.

 

[Helly123]

Yes, sorry, that's better. ΣF=ma along the direction of the string.

But I am confused little bit. Let's say i make SigmaF = ma is
T + mg.cos tetha - centrifugal force = m.a
T + mgcos tetha - mv^2/R = m.a
There's only 1 equation.
To find v and T (2 variables) at least i should have 2 equations..

 

[haruspex]

T + mg.cos tetha - centrifugal force = m.a

You can choose to work in an inertial frame or in the frame of reference of the mass.
In the mass' frame of reference, there is centrifugal force but the acceleration is, by definition, zero.
In the inertial frame, there is no centrifugal force but there is centripetal acceleration, i.e., the component of the resulting acceleration that is normal to the velocity.
The two equations that result are the same.

 

[Helly123]

You can choose to work in an inertial frame or in the frame of reference of the mass.
In the mass' frame of reference, there is centrifugal force but the acceleration is, by definition, zero.
In the inertial frame, there is no centrifugal force but there is centripetal acceleration, i.e., the component of the resulting acceleration that is normal to the velocity.
The two equations that result are the same.

For example i used frame of reference of the mass. SigmaF = m.a
When starts to bend, means T = 0? You said a = 0. Why a zero?
So, W - Centrifugal f = 0
M.g - m.v^2/R = 0
gR = v^2
V = sqrt g(a-b)
V = sqrt g(1/2a)
The right answer is sqrt g(1/3a)
What's wrong??

 

[haruspex]

Why a zero?

In the frame of reference of the mass, the mass is stationary.

M.g - m.v^2/R = 0

What happened to cos theta? You had that in post #42.

 

[Helly123]

In the frame of reference of the mass, the mass is stationary.

What happened to cos theta? You had that in post #42.

Mgcos tetha - m.v^2/R = 0
Costetha gR = v^2
V = sqrt g(a-b) cos tetha
V = sqrt g(1/2a) cos tetha
But still...
The right answer is sqrt g(1/3)

 

[Helly123]

You can choose to work in an inertial frame or in the frame of reference of the mass.
In the mass' frame of reference, there is centrifugal force but the acceleration is, by definition, zero.
In the inertial frame, there is no centrifugal force but there is centripetal acceleration, i.e., the component of the resulting acceleration that is normal to the velocity.
The two equations that result are the same.

For inertial force, there's centripetal force? Can you give me hint how to work using inertial force?

 

[haruspex]

V = sqrt g(a-b) cos tetha

No. When the angle is theta, how far is the mass above the peg? How far below point A does that make it?

For inertial force, there's centripetal force?

Centripetal force is not an actual applied force, like gravity or tension. It is merely that component of the net force which is normal to the velocity. It produces the radial acceleration.
In the vector equation $\Sigma \vec {F_i}=m\vec a$, we can split the acceleration into the tangential acceleration (the component parallel to the velocity vector) and the radial acceleration (the component normal to the velocity): $\Sigma F_i=m\vec {a_t}+m\vec {a_r}$. The centripetal force is $m\vec {a_r}$.

 

[Helly123]

After a week i finally solved this question. Thanks for all your help guys!