TargetIIT

Q1. A small spherical ball of mass m and radius r rolls in a concave surface of radius R with sufficient friction. Find the time period of the resulting SHM.

Q2. A body of mass m and specific gravity d(>1) floats in water. The temperature of both the body and the water is increased by T. The coefficient of linear expansion of the body is α and volume expansion of water is γ .
(i) Find the initial time period.
(ii) What is the new fraction of the body that is immersed in water at equilibrium
(iii) What is the new time period in terms of the original time period?

Asish..

1st one try Energy method :)

:D

mujhe laga hi tha...

thats why i posted the FBD and then deleted it... bcuase i knew everyone will psot energy method... :D

[image]12862092.jpg[/image]

torque about the center of concave .....

Wheres friction :P...

:D

:D usko bhool gaye.......

arey woh lag jaega upar....

par energy method easy hai.....

[b]v=rw
w' = angular velocity of COM of sphere

w'=v/(R-r)= rw/R-r

dw'/dt=r/R-r dw/dt

alpha'=ralpha/R-r

alpha= a/r

a= gsinx/ 1+I/mr[p]2[/p]

a=5gsinx/7

alpha' = 5grsinx/R-r

sinx=x

alpha' = - 5grx/R-r

t=2pi√x/alpha'

T=2pi√(R-r)/5gr


[/b]

yes bhaiya ...
this is wat i did..
at any inst.

mv[p]2[/p]/2 + Iω[p]2[/p]/2 + mg(R-r)(1-cosθ) = const.
==> mω[p]2[/p]r[p]2[/p]/2 + mr[p]2[/p]w[p]2[/p]/5 + mg(R-r)(1-cosθ) = const.
==> mωαr[p]2[/p] + 2ωαmr[p]2[/p]/5 + mgω(R-r)sinθ = 0
==> αr[p]2[/p](7/5) = -g(R-r)θ
==> α = -(5/7)g(R-r)/r[p]2[/p]θ

giving T = 2πr[sqrt]7/5g(R-r)[/sqrt]

but ans is given as 2π[sqrt]7(R-r)/5g[/sqrt]

sry amit ans is T =  2π√[o]7(R-r)/5g[/o]

priyam... energy method me kahaan galti hua hai???

dekhte hain....energy method in SHM bhool gaye hainn ;)

you have made a mistake in

wr

it is actually w(R-r)

[hide]v=ωr
hi hoga..[/hide].

[b]confusion arose bcause there are two ω
one is v=ω[ss]2[/ss]r
another v=ω[ss]1[/ss](R-r)
[/b]
see the fig below...

what is this two ω's and nishant bhaiyya ... y not v=wr .. as ive written wrt COM..

wat abt the second question??

the COM of ball is rotating about center of concave....

and ball is rolling..

and priyam .. y is v=w(R-r) and not v=wr .. is it bcz of the two w'??

[b]as far as i remember... writing this type of energy was discussed long back...
[/b]

do u hav the link...

Q. 1.267 of erodov..

http://targetiit.com/iit_jee_forum/posts/30th_november_2008_762.html

and second waala?

Q.1


KE=(1/2)[(2/5)mr[p]2[/p]+m(R-r)[p]2[/p]]ω[ss]1[/ss][p]2[/p]+(1/2)[(2/5)mr[p]2[/p]+mr[p]2[/p]]ω[ss]2[/ss][p]2[/p]

ω[ss]1[/ss]=v/(R-r)
ω[ss]2[/ss]=v/r

dθ/dt=ω[ss]1[/ss]

and then write everything in terms of a and θ/(R-r)=x
...

[12] well maybe...

[5][5][7][7]     dimaag ke upar se flow kar gayaa....

plz thoda aur explanation de sekte ho.. priyam bhai... but i shudnt ask much .... cuz [hide]ur an emperor and emperors are known to be ruthless [3]

any one volunteer for energy method....[12]

why two omega??
still can't understand!!

http://targetiit.com/iit_jee_forum/posts/30th_november_2008_762.html

[b]isme energy different kyun hai...??[/b]

[i](more energy => less stability)[/i] this reason is not enough for removing an energy....

also... i think this is right... because... answer is matching the force method... But How its correct?

I didnt specify it as a reason for removing energy ...

energy is always translational + rotational

what about that post...

actually in the question it is mentioned that the mass m is small

thus r<<R  which gives that the contribution of velocity by w1 to all the points is almost same ...

as analogous to the case in your link the r is not <<R so the velocity contribution of the w1 there is not same to each point ..

so the case here becomes almost the approaching case of a mass tranlating with v  

But it is not approximation in force method... and answer is same...

but I think i got something... plz see if it can be the reason...

jsut wait for a sec.. don go ofline...

yes abhishek i think thats the point well done :)

finally...

saam se paresaan the ispar.... :)

yes that is indeed a great point to be noticed ...

iN REFERENCE TO # 29


[b][i]here dθ/dt = v/R-r[/i][/b]


FRM WER DUS THIS CUM ??

See post#4 the theta is shown there.
and post#17 w1=v/(R-r)