See both the angles are same......This may not be possible but assume it... (to make calculations easy)....also rod is translating.(assumption)
A pencil is placed vertically with its tip on the edge of a table. It is then given a slight push so that it starts falling towards the ground. It makes collision with ground such that its velocity in vertical direction reduces to zero.
What is its angular velocity just after collision?
* Details and Assumptions *
Length of pencil , the height of table is , Mass per unit length of pencil varies as ,(where, is the distance from the tip of the pencil), take .
(Important note) Assume that when the pencil detaches from table surface It performs pure translational motion.Neglect any change in its com velocity due to this.that is just after it detaches ,assume ( just to make it simpler).
There is sufficient friction on the table. There is sufficient friction on the ground such that just after collision there is no relative motion between the contact surfaces.
Better to solve the question by taking the quantities given in fraction form , as fraction only.
Neglect air drag.
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Centre of mass distance from tip = 2l/3 ; moment of inertia of pencil about com is 1 8 m l 2 ; moment of inertia about tip 2 m l 2
Check when will the normal reaction becomes zero... you will get c o s θ = 2 5 1 6
Velocity at the time of losing contact
Horizontal velocity =128/125(Vx)
Vertical velocity = 72/125
Velocity at the time of collision ; vertical = 144/125(Vy). Horizontal is the same.
Vertical component of velocity reduces to zero(as given)
Ndt = MVy
Similarly , f dt = M (Vx - V)
And the angular impulse equation is
N d t ( 2 l s i n θ / 3 ) + f d t ( 2 l c o s θ / 3 ) = 1 8 m l 2 ω
Since friction will stop the relative motion therefore velocity of bottom most point
V = ω ( 2 l c o s θ / 3 )
Now on solving we get ω = 5 . 0 0