How Can Vector Analysis Improve Understanding of Mechanics Problems?

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coldblood
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Hi friends,
Please help me in solving this problem, I'll appreciate the help.

The problem is as:

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn2/1476555_1461726297387809_897057293_n.jpg

Attempt -

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1526162_1461726404054465_2090696806_n.jpg
https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-prn2/q71/s720x720/1460953_1461726324054473_1703669622_n.jpg
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1512727_1461726390721133_1104335405_n.jpg

Thank you all in advance.
 
on Phys.org
You don't need calculus to solve part 1.
Of the speed, v, with which the bottom of the ladder is moving, what is the component in the direction of the slope of the ladder?
If the top of the ladder is sliding down at speed u, what is its component in that direction?
What is the relationship between those two?
 
If the speed of the bottom point is v, it's component along the ladder be v. cos 300 and if the top end is sliding down with speed u, its component along ladder be u cos600.

And both will be equal hence, v cos300 = u cos600 ∴ u = v√3

But the problem states for the center point.
 
haruspex said:
How will the velocity (vector here) at the mid point relate to the velocities of the endpoints?

hmm! don't know
 
coldblood said:
hmm! don't know
Suppose ra, rb, rm are the position vectors of the two ends and the midpoint. What equation relates them? If you differentiate wrt time you'll get the velocity vectors. What does that equation tell you? (OK, so I said no calculus. Well, not much.)
 
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haruspex said:
Suppose ra, rb, rm are the position vectors of the two ends and the midpoint. What equation relates them? If you differentiate wrt time you'll get the velocity vectors. What does that equation tell you? (OK, so I said no calculus. Well, not much.)

Well, Yes the speed of the mid point is v. Thanks a lot for the good help.
But still sticking for the angular acceleration.
 
coldblood said:
Well, Yes the speed of the mid point is v. Thanks a lot for the good help.
But still sticking for the angular acceleration.
Let the distance of the base of the ladder from the origin be x, and the angle of the ladder be θ at some time t. Write x as a function of θ. Differentiate once, twice, to obtain equations relating the derivatives of θ to v and dv/dt.
Eliminate dθ/dt to obtain an equation for the angular acceleration.
 
haruspex said:
Let the distance of the base of the ladder from the origin be x, and the angle of the ladder be θ at some time t. Write x as a function of θ. Differentiate once, twice, to obtain equations relating the derivatives of θ to v and dv/dt.
Eliminate dθ/dt to obtain an equation for the angular acceleration.

x = l cosθ => differentiating, v = - l sinθ . ω => differentiating, a = - l cosθ . α
∴α = [-a/(l cosθ)]
 
hi haruspex...

I know how to solve the problem ,but I am having a small doubt .If I approach part b) mathematically ,then that gives the correct answer,but when I try to analyse the problem somewhat differently I end up with a different answer.

In part b) if we consider the instantaneous axis of rotation P(changing at every instant)) as marked in the OP ,then the rod is undergoing pure rotation about it.Considering the rotation of bottommost point about P ,if we apply atan = αR ,where atan is the tangential acceleration of the tip and R is the distance between the tip and P .

But atan=dv/dt = 0 ,hence α = 0 .

There is flaw in this reasoning,which I am somehow unable to find .Could you reflect on this ?
 
P is not a fixed axis of rotation. When determining acceleration you can not ignore the motion of the "instantaneous axis" , that the distance from A to P also changes. If you use polar coordinates with respect to P, the tangential acceleration of a point is
atan=2(dr/dt)ω+rα.

Using instantaneous axis is confusing. Better to use the coordinates of the endpoints.
 
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ehild said:
When determining acceleration you can not ignore the motion of the "instantaneous axis" ,
Quite so. An instantaneous axis is fine for angular velocity, but not for angular acceleration. It's one level of differentiation too far. You'd need to take into account the velocity of the axis.

But to be honest I hadn't even noticed the attempt at solving part b in the OP. That has another flaw. It treats the set-up as a ladder falling freely under gravity. That is not what is happening here. We are told that the base of the ladder is not accelerating, for whatever reason. This is a kinematics question, not a kinetics one.
 
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haruspex said:
You must differentiate both sides wrt the same variable, t in this case. What is d/dt (cos θ)?

v = - l sinθ

so, a = - l cosθ. ω
 
ehild said:
P is not a fixed axis of rotation. When determining acceleration you can not ignore the motion of the "instantaneous axis" , that the distance from A to P also changes. If you use polar coordinates with respect to P, the tangential acceleration of a point is
atan=2(dr/dt)ω+rα.

Using instantaneous axis is confusing. Better to use the coordinates of the endpoints.

Thank you very much ehild :smile: ... I would not have been able to figure it out on my own.

But I asked this question for the very same thing which haruspex has addressed.Using the instantaneous axis we can get the angular velocity , but not angular acceleration .

Why ? Could you reflect more on this?

haruspex said:
Quite so. An instantaneous axis is fine for angular velocity, but not for angular acceleration. It's one level of differentiation too far. You'd need to take into account the velocity of the axis.

Thanks haruspex...
 
Tanya Sharma said:
I know how to solve the problem

Tanya please make me understand also.
 
Using the instantaneous axis we can get the angular velocity , but not angular acceleration .

Why ? Could you reflect more on this?

First we have to know: what is instantaneous axis and what is tangential acceleration in this problem. I do not know. Do you?
ehild
 
coldblood said:
v = - l sinθ

so, a = - l cosθ. ω

You are not applying chain rule properly.

x = Lcosθ = f(θ)

dx/dt = df(θ)/dt = [df(θ)/dθ][dθ/dt]
 
ehild said:
First we have to know: what is instantaneous axis and what is tangential acceleration in this problem. I do not know. Do you? ehild

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc3/1471232_1462515570642215_877595177_n.jpg

I think when the rod is at the position AB making angle 450 with horizontal the instantaneous axis would pass through point P and when the rod will slide the new position would be A'B' and instantaneous would shift to P'.
Ans tangential acceleration will be also instantaneous dependent on theses axis of rotation.
 
ehild said:
First we have to know: what is instantaneous axis and what is tangential acceleration in this problem. I do not know. Do you?

ehild

Instantaneous axis is the axis about which the body is in pure rotation.In the present case,it is changing at every moment.

Why I got the idea of using IAOR ? I calculated the angular velocity using IAOR and got the answer .But when I applied it in calculating angular acceleration,it gave incorrect result,hence created the doubt .

By tangential acceleration I meant acceleration of the tip of the rod with respect to the instantaneous axis . It should rather be called linear acceleration .I purposefully used the word tangential so as to avoid confusion in using the condition atan = αR which is quite similar to the rolling constraint 'a = αR' .

If you are hinting me to drop the idea of using the instantaneous axis of rotation ,I am more than happy and willing to accept your advice :smile:
 
Tanya Sharma said:
You are not applying chain rule properly.

x = Lcosθ = f(θ)

dx/dt = df(θ)/dt = [df(θ)/dθ][dθ/dt]

Please see,

x = Lcosθ
dx/dt = v = L(-sinθ. dθ/dt)
v = L(-sinθ. ω)
dv/dt = a = -L(sinθ. α + ω. cosθ . ω)
a = -L(sinθ. α + ω2. cosθ )

now how to replace a ?
 
coldblood said:
Please see,

x = Lcosθ
dx/dt = v = L(-sinθ. dθ/dt)
v = L(-sinθ. ω)
dv/dt = a = -L(sinθ. α + ω. cosθ . ω)
a = -L(sinθ. α + ω2. cosθ )

now how to replace a ?

a = dv/dt = 0 is given in the problem .
 
Tanya Sharma said:
a = dv/dt = 0 is given in the problem .

What about ω?
 
coldblood said:
What about ω?

Use v = -ωLsinθ .

Substitute value of ω in the expression for α .
 
Tanya Sharma said:
Instantaneous axis is the axis about which the body is in pure rotation.In the present case,it is changing at every moment.

And how do you determine its position?


Tanya Sharma said:
If you are hinting me to drop the idea of using the instantaneous axis of rotation ,I am more than happy and willing to accept your advice :smile:

The instantaneous axis is all right in case of rolling, as it is a point of the rolling object, instantaneously in rest. Here it is not part of the rod, it is not fixed, its position depends on the position of the rod...

You can consider the in-plane displacement of a rigid body as a pure rotation about a point O in the plane. Selecting a point P on the body, it displaces by Δr, chord of a circle with radius equal to the distance from the axis. If the displacement is very small, the length of the chord is equal to the length of the arc Δs. If the turning angle is Δθ, the speed is the limit Δs/Δt =RΔθ/Δt v=Rω, with respect to the instantaneous axis of rotation. Now you want the acceleration. For that , you have to take the difference of the velocities at time t1 and at time t2. But the instantaneous axis is not the same at these time instants. The second instantaneous axis has some velocity with respect to the first one, that you have to take into account when determining the change of velocity. Also, the distance R will change... And the tangential direction is different for t1 and t2. It is very complicated . Is not it better to use the fixed Cartesian coordinate system?


ehild
 
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ehild said:
And how do you determine its position?

By drawing perpendicular lines to the direction of motion of any two points on the body .The point of intersection gives you the IAOR .

ehild said:
Is not it better to use the fixed Cartesian coordinate system?

ehild

Yes it is...

Thank you for explaining me why using IAOR can be troublesome .I approached part a) without using calculus ,with some geometrical considerations.When I arrived at the correct result it naturally prompted me to apply the approach in part b) as well .It failed.
 
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I have approached part a) using IAOR and got the correct result .Can you help me understand part a) without using IAOR .

Let rA,rB,rM be the position vectors of the two ends and the midpoint. (A represents top of the rod and B represents bottom tip)

Then rM = (rA + rB)/2

Differentiating , vM = (vA + vB)/2

Now, vB = v (given in the problem)

Is this the correct way to approach ?
 
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(vA-vB).(rB-rA) = 0

This can be rewritten as (vM-vB).(rB-rM) = 0

Is this what I should be getting ?
 
The solution is very simple if you choose the coordinate system so that point A is at (X,0) and point B is at (0,Y). Then the CM is at (X/2, Y/2) and

VCM= 0.5 (dX/dt, dY/dt)

As the length of the rod is constant, X2 + Y2=0, so

X dX/dt+Y dY/dt=0,

dX/dt=V, dY/dt=-(X/Y) V.

X/Y can be written with the angle of inclination, so you get dY/dt in terms of V and the angle.

ehild
 
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