A Leaning Plank. Find where it looses contact.

[ehild]

Let the length of the plank be 2l (for simplicity) and let it make an angle θ with the horizontal when normal force by the vertical wall is 0.

1. Energy Conservation-

ω² = 3g/2l (sinΦ - sinθ), where Φ is the initial angle of the base with horizontal.

Correct so far, but I but I can not follow you after. I write the details to get the angular acceleration

The angular acceleration is obtained by derivation of

ω²=3g/(2L) (sinΦ - sinθ). *

2ω dω/dt=-3g/(2L)(cosθ)ω --->dω/dt=-3g/(4L)cosθ. **

N=0 means that the horizontal acceleration of the centre of mass is zero. The x coordinate of the CM is x_cm=L(cosθ), v_cm=-Lω(sinθ) and

a_cm=-L*(ω²cosθ +dω/dt sinθ ).***

a_cm=0 means that dω/dt =-ω²cotθ, the same you got.

Substituting (*) for ω² and (**) for dω/dt in (***) for the horizontal acceleration, you get the angle when the contact is lost.

I do not understand the signs in your torque method. Show what you consider positive directions. Also, dω/dt is missing from your equation for N2. And you got an expression (*) for ω² already, what are these different new ones?

ehild

 

[zorro]

Substituting (*) for ω² and (**) for dω/dt in (***) for the horizontal acceleration, you get the angle when the contact is lost.

No need of any Langrangian or torque equations?

I do not understand the signs in your torque method. Show what you consider positive directions. Also, dω/dt is missing from your equation for N2. And you got an expression (*) for ω² already, what are these different new ones?

ehild

I took upward direction positive. dw/dt = α in my equations.
I solved for α from the torque equation and Newton's law to find one more expression for α and hence ω.

btw I got the height as 2h/3 not h/3

 

[ehild]

No need of any Langrangian or torque equations?

No need any of them, just get the accelerations dω/dt and a_x by derivation with respect to time.

I took upward direction positive. dw/dt = α in my equations.
I solved for α from the torque equation and Newton's law to find one more expression for α and hence ω.

Check the signs in your equations. But the torque method is quite complicated.

btw I got the height as 2h/3 not h/3

2h/3 is correct. Great work!

First I wrote up the equations of motion for the translation of the CM and for the rotation around it. It was quite complicated. Then I used the Lagrangian to find the equation of motion in terms of the angle. But I had to solve a differential equation to get the angular speed. The easiest way was using conservation of energy.

ehild

 

[ashishsinghal]

Thanks to all for this wonderful effort. I am really thankful.Never could have possibly got to that without your help.
Special thanks to ehild.