How Do You Solve Complex Pendulum Equations in Physics Homework?

[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!