Question

A uniform plank of mass M and length L leans against a vertical wall. The initial angle it makes with the horizontal ground is 60º. It is let go from rest and slides down frictionlessly. The moment it leaves contact with the wall, what angle does it make with the ground? There is no friction anywhere.

Answer 1

For simplicity, I will use ∅ as the angle between the ladder and the floor, with ∅ at the beginning equal to 60 degrees.

Considering rotation of the ladder with respect to the instantaneous centre of rotation S,

Let angular acceleration is a.

Let T is torque. Moment of Inertia is I = ml²/3

I*a= T
ml²a/3 = mglcos(∅)/2
a = 3gcos(∅)/2l

Also,
ml²a/3 = mglcos(∅)/2
Multiplying the above equation by angular velocity w, we get

ml²a*w/3 - mglcos{∅}*w/2 = 0
This can be written as :

Writing a=dw/dt and integrate it

Integration of 0 is c (constant)

d/dt(ml²*w²/6 + mglsin(∅)/2) = 0

ml²w²/6 + mglsin(∅)/2 = C
(where C is a constant)

At time t=0, w = 0 ∅=60
Thus, C = mglsin(60)/2

Therefore, ml²w²/6 + mglsin{∅}/2 = mglsin{60}/2
w² =3g/l({sin∅ - sin60})

Let the x co-ordinate of the ladder's center of mass be X.
X = l/2cos{∅}

When the ladder loses contact with the wall, horizontal force becomes zero.
Therefore, X'' = 0

differentiate x two times and

Put d∅/dt = w and dw/dt = a

X'' = l/2(cos{∅}*w²+ sin{0}*a) = 0

cos{∅}*w² = -sin{∅}*a

Substituting the values of w² and a that we found earlier, we get:
sin{∅} = 2/3sin{60}
∅ = arcsin{(2sin{60}/3)}

∅ = 35.26
Thus at 35.26 degrees angle with the floor, the ladder loses contact.