33just neglect gravity here because as per my experience the spring problems in which the setup can't perform SHM under the effect of gravity considering spring to be removed... ans is free of gravity i.e you can safely take g=0.
1in equlibrium ,
When further displaced by x ,
E = 12mv2 + 12Iω2 + 12k( x + x0)2 - mgx sinθ
since for S.H.M → E = constant
hence dEdt = 0
→ 0=mv\frac{dv}{dt} + I\omega \frac{d\omega }{dt}+k(x+x_{0})\frac{dx}{dt}-mgsin\theta \frac{dx}{dt}
Substituting dvdt = a , ω = v/R , I = 12mR2 , dωdt = α = aR , dx/dt = v and kx0 = mgsinθ
we get ,
3ma = -2kx .............................(1)
Since f=\frac{1}{2\pi }\sqrt{\left|\frac{a}{x} \right|}
from (1) we get
f = f=\frac{1}{2\pi }\sqrt{\frac{2k}{3m} }
now substitute the given values as in the question u will get f = 0.56 Hz [1]
1
Manmay kumar Mohanty
·2010-04-28 21:11:48
why yet no confirmation ??
aditya babua please confirm na
bas atta ho question type kar deta HAI AUR chala jata hai...[3]
1cant we to it by writing 2 equations 1 linear i.e including force due to mass, friction and spring force... and that ffriction.R= I∞
the finding a in terms of -x.... and the coefficient of x will be omega2....
??????
1
varun.tinkle
·2010-05-01 21:28:32
u can do it with the normal force method when the com is displaced by a distance x then
force acting is k(xequilibrium+x)-mgsin@=ma
but Kxeq=mgsin@
therfore force acting is kx
not for rotation forces
fr=1/2mr^2 @
now we know @
now since it is not slipping v=Rw
i think it is enuf hints to solve the sum
Setermine the natural freuency of vibration of 100N disc , assume the disc does not slip on the incline surface..