A plank of length 2L leans against a frictionless wall at a height h.
It starts to slip downward without any friction. This will happen for some time, but suddenly the plank will lose contact with the wall, and fall to the ground all of a sudden! You can try to observe this at home too.
The problem is to calculate the height at which the plank loses contact with the wall.

- Mani_Jha

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- Mani_Jha

|dw:1332297485832:dw|
I think this is a really tough problem. The book from which I have taken says that 'it is a tricky question which needs you to use all the concepts you've learned in rotational mechanics so far'
However, this sum is very good for practicing rotational mechanics. Think on this, it's not a problem if you can't do this, but if you don't even try, it is!
Hint: Try to analyze the motion of the center of mass.

- AravindG

good question mani

- AravindG

actually i hav dpne a very similiar problem

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## More answers

- AravindG

which book u got this qn?

- Mani_Jha

An Introduction to Mechanics by Daniel Kleppner

- anonymous

@JamesJ @nikvist @Jemurray3

- anonymous

I will be back

- Mani_Jha

They ain't members of this group. Will they come?

- AravindG

james has been away for sometime

- anonymous

No harm in trying ;-)

- AravindG

agreed

- anonymous

|dw:1332332991816:dw|Forces acting on the Plank must be \(N_{ground}, N_{wall}\) and \(W_{gravity}\) or \(W_{g}\).

- anonymous

|dw:1332333331620:dw|Rotational axis passes from B, Can you think of a better location for Rotational axis?

- Mani_Jha

Rotational axis is about the center of mass. The motion is a rotation about the center of mass axis.

- anonymous

I won't be able to eliminate the Normal reaction from Wall, if I chose this. Hmm I guess that's not bad, is it? because the the plank must leave the wall when N (the normal reaction from wall) \(\to\) Zero. So, I do need the Normal reaction form wall.

- Mani_Jha

Here's what I've thought so far:
|dw:1332329477829:dw|
The motion of the plank is a rotation about the center of mass axis, an axis perpendicular to the plane of the computer screen. There are two forces, and thus two torques that cause this rotation â€“ one due to N1, the normal force due to the wall, and the other due to N2, the normal force due to the floor.
Torque due to N1=N1Lsin(theta)
As the plank falls to the ground, theta decreases, and so must the torque due to N1(This somewhat gives the idea that contact will be eventually lost). This torque acts in a direction into the plane of the computer screen.
Torque due to N2=N2Lcos(theta)
As the plank falls to the ground, this torque increases(costheta increases). This torque also acts into the plane of the computer screen.
I am finding it difficult to analyze the motion of the center of mass. Will it follow a straight vertical path or a curved one?

- anonymous

Moment of Inertia of the Rod about the point B must be \(\frac{1}{3} m\times 4 \times l^2\).
\(\frac{4}{3} m l^2\times \alpha = mg*l\cos \theta - N_{wall}l\sin \theta\), where \(\alpha\) is the angular acceleration.

- anonymous

A Curved Path, it would have been straight line fall if the wall wasn't there.

- Mani_Jha

Oh, but the axis of rotation passes through the center of mass, not B.

- anonymous

For the translation motion, \(N_{wall} = \sum F \implies N_{wall} = m*a\).
The platform is stationary or the ground is stationary i.e not moving lol... \(a = \alpha l \)
Oh, a typo... the equation of rotational motion about B must be
\[\frac{4}{3} m l^2\times \alpha = mg*l\cos \theta - N_{wall}2l\sin \theta\]

- anonymous

Mani it doesn't matter from where the rotational axis passes, note that I did account for the change in the Moment of Inertia.

- Mani_Jha

Oh, ok.
We could put N=0 in that equation, and then we would get a differential equation...can you solve that?

- anonymous

great question pal. i'll be back

- anonymous

How would you get an expression for a, the linear acceleration? I need to solve it on the paper first.

- anonymous

JamesJ came online, he might get here now...

- Mani_Jha

We can try putting: \[a=alphal\] in that equation for the rotation.
And write costheta=base/hypotenuse
=\[\sqrt{4l ^{2}-h ^{2}}/2l\]
where h is the height(variable)
That would be a differential equation to solve........

- anonymous

\(a = \alpha l\) is valid as long as the plank doesn't lose contact with wall, after the plank losses contact, \(a \to 0\)... after losing the contact the plank must only have the the angular acceleration.

- anonymous

\[\frac{4}{3} m l^2\times \alpha = mg*l\cos \theta - N_{wall}2l\sin \theta\]\[\implies \frac{4}{3} m l^2\times a \times \frac1{l} = mg*l\cos \theta - N_{wall}2l\sin \theta\]\[\implies \frac{4}{3} m l\times a= mg*l\cos \theta - N_{wall}2l\sin \theta\]

- anonymous

\[\implies \frac{4}{3} a= g*\cos \theta - 2a \sin \theta\]I think I did some mistake somewhere, because now if I take a -> 0, I will get \(g\cos \theta =0\) :(

- Mani_Jha

a is the acceleration of the part of the plank in contact with the wall, right? So, when contact is lost, a must become equal to g, isn't it? Because it's falling freely under gravity then.

- anonymous

What's the answer btw?

- Mani_Jha

2h/3

- Mani_Jha

It is an interesting problem, but trying to solve it gives me headaches!

- anonymous

Then I did something wrong, because I didn't account for h at any time in my solution :/ I won't get answer in terms of h but in terms \(l\).

- anonymous

It's not that hard, the motion is simple... I am messing up somewhere :/

- anonymous

I will try to do it again.

- anonymous

Oh I know what I did, I assumed the vertical linear acceleration to be Zero, biggest mistake ever!!! and stupid too :/

- Mani_Jha

Hey Ishaan, will the total mechanical energy of the centre of mass of the system be conserved? If so, the initial energy of the centre of mass is
\[mgl \sin \theta\]
The final energy would be :
\[mv ^{2}/2+Iw ^{2}/2\]
where
\[v=\sqrt{v(x)^{2}+v(y)^{2}}\]
v(x) and v(y) are the horizontal and vertical displacements of the center of mass respectively.
In terms of energy, I think the problem would become easier.
I know that the constraint relation between the vertical displacement of the top of the plank y and horizontal displacement of x of the bottom of the plank is:
\[y=x \tan \theta\]
The vertical and horizontal displacements of the center of mass is somehow proportional to y and x. I need to find this relation.

- anonymous

Yes, Mani... mechanical energy is conserved.

- anonymous

mechanical energy as well as angular momentum also is conserved here due to external force is present here...

- AravindG

wow!! i told u i did a similiar problem!!but the most exciting part is that both the answers are same!!!let me post that question ,,you would realize why the answers are same while doing it

- AravindG

thx for posting quality questions here Mani!!

- anonymous

u guys have ncert 11th text book look under it in examples for the wonderful explanation to this question:under rotational motion:EXAMPLE 7.9

- AravindG

2h/3

- Mani_Jha

@Taufique, Angular momentum about which point is conserved? There is a net external torque in this case, so how can the system's angular momentum be conserved?

- anonymous

@mani here normal force and gravitational force both are internal force and torque by these force also be internal not external torque.Hence system,s angular momentum be conserved..

- anonymous

It's possible without friction, no one said you to consider Plank before the motion, just do what the question says.

- anonymous

yes ishaan u were right i did not read the question properly
|dw:1332582251137:dw|
note above that the rod sides because of normal force N
no other forces in the left or right direction so when this loses contact,this N=0(so now the rod is only under gravity influence and falls abruptly down)
also when F=0 torque due to F is no longer there and is 0
taking my origin at point where N' acts,
torque=0=2l*F*sin@
we can put any value for sin@ including 0 so i havent put a quantitative explanation here however this is my idea see whether u can manipulate it
by getting sin@ we can easily find the height

- anonymous

Soo... I finally have the solution. It's long, really long with loads of equations. Let me type it out for you guys.

- anonymous

Length of the plank is \(2l\), the initial angle is \(\theta_0\), initial height from the ground to the point of contact is \( h\).
\(h = 2l\sin \theta_0 \tag1\)
|dw:1332593176429:dw|
I think we all know the motion of center of mass, a curved path.
Now, let the angle at which the plank loses contact be \(\theta\). Moment of Inertia about the center of the mass is, \(\frac{1}{12} m (2l)^2 = \frac{1}3 m l^2\).
Rotational Motion's equation, \(\sum \tau = I \alpha \).
\[ \frac1 3 ml^2 \alpha = N_g l \cos \theta - 2N_w l \sin \theta \]\(N_g\) is the normal reaction from the ground, and \(N_w\) is reaction from the wall, \(\alpha \) is the angular acceleration.
For the plank to lose contact from the wall, \(N_w \to 0 \).
\[\implies \frac 1 3 ml^2 = N_g l \cos \theta \]
Translational Motion's equations.
\[N_w = ma_x \implies a_x = 0\tag2\]\[N_g - W = ma_y \implies N_g = W + ma_y \tag3\]Note. \(W\) is the gravitational pull on the plank.
At any point of time \((x_{cm}, y_{cm}) = (l \cos \theta, l\sin \theta)\)
\(x = l \cos \theta \) after double differentiation w.r.t \(t\) (time), \(\frac{d^2x}{dt^2} = - l \cos \theta \left(\frac{d\theta}{d t}\right)^2 - l\sin \theta \frac{d^2 \theta}{dt^2} \).
Using equation (2).
\[-\omega \frac{\cos \theta}{\sin \theta} = \alpha\tag4\]\[\text{Note. } \left(\frac{d^2 \theta}{dt^2}, \frac{d\theta}{dt}, \frac{d^2y}{dt^2}, \frac{d^2x}{dt^2} \right) =\left(\alpha, \omega, a_y, a_x \right)\]Similarly, \(a_y = -l \sin \theta\left(\frac{d\theta}{dt}\right)^2 + l\cos \theta \frac{d^2\theta}{dt^2} \tag5\)
From the energy theorem, \[mgl\sin \theta_0 - mgl\sin \theta = \frac1 2 m v^2 + \frac 1 2 I \omega^2 \tag6\]\(v = \omega l \tag 7\)
Using (7) and (6), \(mgl(\sin \theta_0 - \sin \theta ) = \frac 2 3 m \omega^2 l^2\tag 8\)
From equation (3), (5) and \(\frac{1}{3}ml^2 = N_g l\cos \theta\).
\[ \frac 3 4 g\sin \theta = l \omega^2 \tag9 \]
Using equations (9) and (8),
\[g(\sin \theta_0 - \sin \theta ) = \frac{2}{3} \times \frac 3 4 g \sin \theta \]\[\implies \sin \theta_0 \frac 2 3 = \sin \theta \implies 2l \frac2 3 \sin \theta_0 = 2 l \sin \theta \]From (1), \[\frac 2 3 h = h_f\] \(h_f\) is the height where the plank loses the contact from the wall.

- anonymous

I haven't shown all the work, I think you can do that on your own.

- Mani_Jha

Great work, Ishaan. You rock!

- anonymous

Thanks :-)

- anonymous

|dw:1332734203601:dw|
|dw:1332734393976:dw|
solution based on conservation of energy::
Let the verticle position from the ground
where leans losses the contact of the wall be
y.
From the conservation of energy::
mg(h/2)(at initial)=(mgY/2+1/2I(omega)^2)(just
before loss the contact of the wall)....(i)
at a time of loss the contact:
mg(y/2)+1/2I(omega)^2=mg(y/2)+1/2I'(omega')^2
Find the relation between omega before and
after loss the contact.
put I=(4ml^2)/12,I'=(4ml^2)/3...
you will got...(omega)^2=4*(omega')^2....(ii)
now Let the finalangular velocity omega'' at
just befor come on the ground.
then mgy/2=1/2I'(omega'')^2
=>(omega'')^2=3/4(mgy/l^2)...(iii)
since torque=I'(angular acceleration)
mglcos(theta)=(4ml^2)/3*(angular
acceleration)
therefore angular acceleration=(3g/4l)cos
(theta)...(iv)
apply this formula.
(omega'')^2=(omega')^2+2(angular acceleratio)
*(theta)
since (theta is very very small the tan
(theta)=(theta))
=>(omega'')^2=(omega')^2+2(angular
acceleratio)*(tan(theta))
=>(omega'')^2=(omega')^2+2((3g/(4l))cos
(theta))*(tan(theta))
=>(omega'')^2=(omega')^2+2((3g/4)sin(theta))
=>(omega'')^2=(omega')^2+2((3g/4l)*y/(4l))
....find (omega')^2 from this equation and
eqn(iii)..
i.e (omega')^2= 3gy/(8l^2)
hence (omega)^2=4*3gy/(8l^2)=3gy/2l^2...from
(ii)
....after putting this value in eqn (i) the we
get:
=>mg(h/2)=mg(y/2)+1/2(4ml^2/12)*(3gy/2l^2)
=>h/2=y/2+y/4
=>h/2=3y/4
=>y=2h/3....this is that verticle position
from the ground where leans losses the contact
of the wall.

- anonymous

@Ishaan94 Nice work :) I'm sorry I've been away for a long time, I've just been busier than expected lately, but you seem to have things under control. Hopefully I'll be around more in the near future.

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