# Rotational Motion Problem - 1

1. Dec 16, 2013

### coldblood

Hi friends,

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

2. Dec 16, 2013

### haruspex

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?

3. Dec 17, 2013

### coldblood

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.

4. Dec 17, 2013

### haruspex

How will the velocity (vector here) at the mid point relate to the velocities of the endpoints?

5. Dec 17, 2013

### coldblood

hmm! don't know

6. Dec 17, 2013

### haruspex

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.)

7. Dec 17, 2013

### coldblood

Well, Yes the speed of the mid point is v. Thanks a lot for the good help.
But still sticking for the angular acceleration.

8. Dec 17, 2013

### haruspex

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.

9. Dec 17, 2013

### coldblood

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

10. Dec 17, 2013

### Tanya Sharma

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 ?

11. Dec 18, 2013

### ehild

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.

12. Dec 18, 2013

### haruspex

You must differentiate both sides wrt the same variable, t in this case. What is d/dt (cos θ)?

13. Dec 18, 2013

### haruspex

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.

14. Dec 18, 2013

### coldblood

v = - l sinθ

so, a = - l cosθ. ω

15. Dec 18, 2013

### Tanya Sharma

Thank you very much ehild ... 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?

Thanks haruspex...

16. Dec 18, 2013

### coldblood

Tanya please make me understand also.

17. Dec 18, 2013

### ehild

First we have to know: what is instantaneous axis and what is tangential acceleration in this problem. I do not know. Do you?

ehild

18. Dec 18, 2013

### Tanya Sharma

You are not applying chain rule properly.

x = Lcosθ = f(θ)

dx/dt = df(θ)/dt = [df(θ)/dθ][dθ/dt]

19. Dec 18, 2013

### coldblood

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.

20. Dec 18, 2013

### Tanya Sharma

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