Does Really Energy is Conserved..??

A Mettalic disc of radius R is rotating about its axis with constant angular velocity ω \omega Then Find The Potential difference develop between the centre and Circumference of disc.

DETAILS AND ASSUMPTIONS:

m e = 9.1 × 10 31 k g q e = 1.6 × 10 19 c R = 4 m ω = 1.76 r a d / s { { m }_{ e }=9.1\times { 10 }^{ -31 }kg\\ { q }_{ e }=1.6\times { 10 }^{ -19 }c\\ R=4m\\ \omega =\sqrt { 1.76 } rad/s\\ } .

NOTE: MULTIPLY YOUR ANSWER BY 10 11 { 10 }^{ 11 } .


The answer is 8.

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3 solutions

Deepanshu Gupta
Oct 11, 2014

C o n c e p t Concept :

No energy is not Conserved since For a short time There is Current is flowing in Metallic disc when Free electrons are moving with respect to metal . Finally they get stops moving with respect to metal. So current will also becomes Zero since it's duration is very small so there is no significant energy loss.

And it is understood That this rod is Rotating By an external agent so that it can move with constant angular velocity , Otherwise Rod will never be rotated with constant angular velocity.

N o t e Note :

Energy is loss when current is flowing is explained by
" J o u l e s Joules L a w Law o f of H e a t i n g Heating "

S o l u t i o n Solution :

It is very easy by equating Pseudo Force on electron to the Electric Force on the electron After The Equilibrium is set-up with respect to metal disc.

Let

System (electron)

&

Reference frame( metallic disc)

F n e t = m ω 2 r e E = m ω 2 r E = m ω 2 r e \E m o t i o n a l e m f = 0 R E . d r cos 0 = 0 R m ω 2 r e . d r = m e ω 2 R 2 2 e { F }_{ net }=m{ \omega }^{ 2 }r\quad \\ eE=m{ \omega }^{ 2 }r\\ E=\frac { m{ \omega }^{ 2 }r }{ e } \\ { \E }_{ motional\quad emf }=\quad \int _{ 0 }^{ R }{ E.dr\cos { 0 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad =\quad \int _{ 0 }^{ R }{ \frac { m{ \omega }^{ 2 }r }{ e } .dr } \\ \quad \quad \quad \quad \quad \quad \quad \quad =\frac { { m }_{ e }{ \omega }^{ 2 }{ R }^{ 2 } }{ 2e } .

Bro, in a conducting disc, there is no resistance (ideal case) so energy is not lost due to joules law of heating, your method is absolutely correct, i too did the same way,but being a conductor, the electrons instantenously acquire the stable configuration , and no energy loss takes place (in principle)

Mvs Saketh - 6 years, 3 months ago

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I agreed ! Sorry here I Loosely stated that it is conducting Disc . Instead of this I should state that it is "Mettalic Disc' . My Intention was to say that , disc has free electron's so that they can move , But Yes you are correct It should not be conductor .

Thanks I edited Acordingly!

Deepanshu Gupta - 6 years, 3 months ago

At last line. It looked like (emotional emf) lol

Md Zuhair - 2 years, 4 months ago

R o t a t i o n a l k i n e t i c e n e r g y w i l l b e e q u a l t o V q I w 2 2 = V q u s i n g t h e g i v e n v a l u e s , V = 8.008 Rotational\quad kinetic\quad energy\quad will\quad be\quad equal\quad to\quad Vq\quad \\ \frac { I{ w }^{ 2 } }{ 2 } =Vq\\ using\quad the\quad given\quad values,\\ V=8.008

This is not correct solution Since Energy is not conserved..!! You Should See My Solution.

Deepanshu Gupta - 6 years, 8 months ago

Friend sorry to say that but here your method as well as your answer is wrong.Here the P.D does not provide the rotational energy. Here P.D develops because an electric field came into exsistance inside the disc due to the the unequal distribution of electrons. This is because , when the this is rotated ,their is no force acting on the free of metallic disc initially. Thus they are acted upon a centrifugal forces which is outward.And due to this only the electron;s conc. on the rim increases And for this; E = m ( w ) 2 R E=m*(w)^2*R . (for electrons on rim) and V = E R V=E*R . And by this the p.d comes to be =16V. THANK YOU.

A Former Brilliant Member - 6 years, 8 months ago

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though your way is right answer is wrong m v 2 / r = f o r c e v = r w mv^2/r=force v=rw . therefore m ( r 2 ) ( w 2 ) / r = q E ( f = q E ) m r w 2 = q e R = V m w 2 / q = E m(r^2)(w^2)/r=qE(f=qE) m*r*w^2=q*e*R=V m*w^2/q=E . which comes out to be 10 V 10V .

Rohan Leekha - 6 years, 8 months ago

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