Pens Have Other Uses Too

Consider the pen to be a thin rod having uniform mass distribution over its length

Let $m =$ Total mass of pen

$l =$ length of pen $= 15 cm$

$\omega =$ angular velocity of pen when it hits the table

Initially when the pen is held vertical, the center of mass of pen is $\frac{l}{2} cm$ above the level of table and when it is released, it rotates about it's bottom tip and hits the table by which the height of center of mass of pen becomes zero w.r.t table

Let $y_{1} =$ Initial height of the center of mass w.r.t table $= \frac{l}{2} cm$

$y_{2} =$ Final height of the center of mass w.r.t table $= 0 cm$

Hence,the change in gravitational potential energy of pen is responsible for the rotational motion of pen about its lower tip.

$\therefore K = \frac{1}{2}I\omega^{2} =$ Rotational Kinetic energy of the pen

Where, $I = \frac{ml^{2}}{3} =$ Moment of Inertia of pen about its bottom tip

Hence, by using work energy theorem,

(change in gravitational potential energy of pen) + (Rotational Kinetic energy of the pen) = 0

$\therefore mg(y_{2} - y_{1}) + \frac{1}{2}I\omega^{2} = 0$ where $g =$ Gravitational acceleration = $9.8 ms^{-1}$

$\therefore mg(y_{1} - y_{2}) = \frac{1}{2}I\omega^{2}$

$\therefore mg( \frac{l}{2} - 0) = \frac{1}{2}\frac{ml^{2}}{3}\omega^{2}$

$\therefore \omega = \sqrt{\frac{3g}{l}}$

Let $v =$ velocity of upper end of the pen when it hits the ground $= \omega l$ $= \sqrt{\frac{3g}{l}} l = \sqrt{3gl}$

$= \sqrt{3 \times 9.8 \times 15 \times 10^{-2}} ms^{-1} = 2.1ms^{-1}$

simple..........g=ra (a=angular acceleration, acceleration due to gravity,r=.15) ,find a,then w * w=2 * a pi/4 (w =angular velocity,pi/4 =angular displacement) find w,then v=w r (velocity)

BY THE CONSERVATION OF ENERGY mgl\2 = 1\2(I square of angular velocity ). THEN APPLY v = lw

apply work energy theorem considering the heigth of cm wrt table. mgl/2=1/2Iw^2. w=rt(3gl) putting all the values w =2.1