physics_doubth_questions Physics doubt questions
 Q-1. A flywheel in form of disc is rotating about an axis passing thru the center and perpendicular to the plane looses 100J of energy when slowing down from 60 rpm to 30 rpm. Find the moment of inertia about the same axis and change in angular momentum. Answer-1:𝜔1=60 rpm=2𝜋 rad/sec${𝜔}_{1}^{}=60\phantom{\rule{0.22em}{0ex}}rpm=2𝜋\phantom{\rule{0.22em}{0ex}}rad/sec$𝜔2=30 rpm=𝜋 rad/sec${𝜔}_{2}^{}=30\phantom{\rule{0.22em}{0ex}}rpm=𝜋\phantom{\rule{0.22em}{0ex}}rad/sec$Loss in energy = 100 J Initial kinetic energy =K1=12I𝜔21=12I ((2𝜋))2 $={K}_{1}=\frac{1}{2}I{𝜔}_{1}^{2}=\frac{1}{2}I\phantom{\rule{0.22em}{0ex}}\left(2𝜋{\right)}_{}^{2}\phantom{\rule{0.22em}{0ex}}$ Final kinetic energy =K2=12I𝜔22=12I ((𝜋))2 $={K}_{2}=\frac{1}{2}I{𝜔}_{2}^{2}=\frac{1}{2}I\phantom{\rule{0.22em}{0ex}}\left(𝜋{\right)}_{}^{2}\phantom{\rule{0.22em}{0ex}}$ Loss in kinetic energy =K1-K2=12I [((2𝜋))2 -((𝜋))2]$={K}_{1}-{K}_{2}=\frac{1}{2}I\phantom{\rule{0.22em}{0ex}}\left[\left(2𝜋{\right)}_{}^{2}\phantom{\rule{0.22em}{0ex}}-\left(𝜋{\right)}^{2}\right]$ 12I [((2𝜋))2 -((𝜋))2]=100$\frac{1}{2}I\phantom{\rule{0.22em}{0ex}}\left[\left(2𝜋{\right)}_{}^{2}\phantom{\rule{0.22em}{0ex}}-\left(𝜋{\right)}^{2}\right]=100$⟹I=2003𝜋2=6.754 kg m2 $⟹I=\frac{200}{3{𝜋}^{2}}=6.754\phantom{\rule{0.22em}{0ex}}kg\phantom{\rule{0.22em}{0ex}}{m}^{2}\phantom{\rule{0.22em}{0ex}}$Change in momentum=I((𝜔1-𝜔2))=2003𝜋2×((2𝜋-𝜋)) =21.22 kg m2/sec $=I\left({𝜔}_{1}-{𝜔}_{2}\right)=\frac{200}{3{𝜋}^{2}}×\left(2𝜋-𝜋\right)\phantom{\rule{0.22em}{0ex}}=21.22\phantom{\rule{0.22em}{0ex}}kg\phantom{\rule{0.22em}{0ex}}{m}^{2}/sec\phantom{\rule{0.22em}{0ex}}$ Q-2. Two wheels of moment of inertia 4 kgm2 $4\phantom{\rule{0.22em}{0ex}}kg{m}^{2}\phantom{\rule{0.22em}{0ex}}$ rorate side by side at the rate of 120 rev/min and 240 rev/min respectively. in opposite directions. If now both the wheels are coupled by means of weightless shaft so that bothe the wheels now rotate with common angular speed. Find the new speed of rotation. Answer-2:𝜔1=120 rpm=4𝜋 rad/sec${𝜔}_{1}^{}=120\phantom{\rule{0.22em}{0ex}}rpm=4𝜋\phantom{\rule{0.22em}{0ex}}rad/sec$𝜔2=240 rpm=8𝜋 rad/sec${𝜔}_{2}^{}=240\phantom{\rule{0.22em}{0ex}}rpm=8𝜋\phantom{\rule{0.22em}{0ex}}rad/sec$ in opposite directionLet common angular speed=𝜔$Let\phantom{\rule{0.22em}{0ex}}common\phantom{\rule{0.22em}{0ex}}angular\phantom{\rule{0.22em}{0ex}}speed=𝜔$ Total Initial momentum=L=4×8𝜋-4×4𝜋=16 𝜋 kg m2/sec$=L=4×8𝜋-4×4𝜋=16\phantom{\rule{0.22em}{0ex}}𝜋\phantom{\rule{0.22em}{0ex}}kg\phantom{\rule{0.22em}{0ex}}{m}^{2}/sec$ When wheels are joined the moment of inertia = 4 + 4=8 kgm2 $=8\phantom{\rule{0.22em}{0ex}}kg{m}^{2}\phantom{\rule{0.22em}{0ex}}$ a8 𝜔=16 𝜋 ⟹𝜔=2𝜋 rad/sec = 60 rpm $\begin{array}{l}8\phantom{\rule{0.22em}{0ex}}𝜔=16\phantom{\rule{0.22em}{0ex}}𝜋\phantom{\rule{0.44em}{0ex}}\\ ⟹𝜔=2𝜋\phantom{\rule{0.22em}{0ex}}rad/sec\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}60\phantom{\rule{0.22em}{0ex}}rpm\phantom{\rule{0.22em}{0ex}}\end{array}$