A uniform thin wooden plank AB of length L an | vedclass

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(C) When an impulse $J$ is given to the wooden plank $AB$,there are two extreme possibilities:$(i)$ The plank rotates about $A$ as soon as the impulse

Vedclass · Accepted Answer

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

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(C) When an impulse $J$ is given to the wooden plank $AB$,there are two extreme possibilities:$(i)$ The plank rotates about $A$ as soon as the impulse is imparted.$(ii)$ The plank moves vertically upward without any rotation.Case $I$: Rotation about $A$.Angular momentum about $A$ is conserved: $I_{A} \omega = J L$where $I_{A} = \frac{M L^{2}}{3}$ is the moment of inertia about $A$.$\frac{M L^{2}}{3} \omega = L J \Rightarrow \omega = \frac{3 J}{M L}$The linear velocity of the centre of mass is $v = \left(\frac{L}{2}\right) \omega = \frac{3 J}{2 M}$.Using energy conservation,$\frac{1}{2} M v^{2} = M g h \Rightarrow h = \frac{v^{2}}{2 g} = \frac{9 J^{2}}{8 M^{2} g}$.Case $II$: Pure vertical motion without rotation.By conservation of linear momentum,$J = M v \Rightarrow v = \frac{J}{M}$.Using energy conservation,$\frac{1}{2} M v^{2} = M g h \Rightarrow h = \frac{v^{2}}{2 g} = \frac{J^{2}}{2 M^{2} g}$.Since the actual motion involves a combination of rotation and translation,the height $h$ attained by the centre of mass lies between these two extreme cases:$\frac{J^{2}}{2 M^{2} g} < h < \frac{9 J^{2}}{8 M^{2} g}$.

Column-$I$	Column-$II$
$(1)$ $SI$ unit of torque	$(a)$ $m$
$(2)$ $SI$ unit of radius of gyration	$(b)$ $N\,m$
	$(c)$ $Js^{-2}$

$A$ uniform thin wooden plank $AB$ 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 the centre of mass rising a distance $h$ from the surface of the table. Then,

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ $SI$ unit of torque $(a)$ $m$

$(2)$ $SI$ unit of radius of gyration $(b)$ $N\,m$

$(c)$ $Js^{-2}$

$A$ particle performs uniform circular motion with an angular momentum $L$. If the angular frequency of the particle is doubled and its kinetic energy is halved,what will be its new angular momentum?

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third,the new angular momentum becomes:

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