A uniform thin wooden plank A B of length L a | vedclass

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Question

$(c)$ When an impulse $J$ is given to wooden plank $A B$, then there are two possibilities

$(i)$ Plank started to rotate about $A$ as soon as impul

Vedclass · Accepted Answer

$J^{2} / 2 M^{2} g < h < 9 J^{2} / 8 M^{2} g$

Vedclass · Answer

$(c)$ When an impulse $J$ is given to wooden plank $A B$, then there are two possibilities

$(i)$ Plank started to rotate about $A$ as soon as impulse is imparted.

$(ii)$ Plank moves vertically without rotation.

Now, we consider both cases one by one.

Case $I$

In case rotation occurs about $A$, angular momentum about $A$ is conserved. $\Rightarrow \quad I_{A} \omega=J L$

where, $I_{A}=$ moment of inertia of plank about $A=\frac{M L^{2}}{3}$

and $\omega=$ angular speed of plank about $A$.

$\frac{M L^{2}}{3} \omega=LJ$

$\Rightarrow \quad \omega=\frac{3 J}{M L}$

So, linear velocity of centre of mass,

$v=\left(\frac{L}{2}ight) 	imes \omega \Rightarrow v=\frac{3 \cdot J}{2 M}$

If $h=$ maximum height attained by centre of mass, then

By energy conservation, we have

$\frac{1}{2} m v^{2}=m g h \Rightarrow h=\frac{v^{2}}{2 g}=\frac{9 J^{2}}{8 M^{2} g}$

Case $II$ If there is no rotation of plank, then by conservation of linear momentum, we have

and from conservation of energy, we have

$\frac{1}{2} m v^{2}=m g h \Rightarrow h=\frac{J^{2}}{2 M^{2} g}$

Clearly, maximum height attained by centre of mass of plank is

$\frac{J^{2}}{2 M^{2} g} < h < \frac{9 J^{2}}{8 M^{2} g}$

A uniform thin wooden plank $A B$ of length $L$ and mass $M$ is kept on a table with its $B$ end slightly outside the edge of the table. When an impulse $J$ is given to the end $B$, the plank moves up with centre of mass rising a distance $h$ from the surface of the table. Then,

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