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5 Apr 2008 11:23:23 IST
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2gh should be equal to minimum velocity for a mass to complete a vertical loop when in horizontal position = 
2gR
6 Apr 2008 10:05:57 IST
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62)
a)
the work done in projecting the block up the incline=mglsin@
the work done in projecting up the sphere=change in potentiol energy of the system=mgr(1-cos@) from the geometry of the figure
1/2mv^2=mgr(1-cos@)+mglsin@
solving it
v= root 2g(r(1-cos@) +lsin@)
b)
1/2m(2v)^2=mgr(1-cos@)+mglsin@+1/2m(v1)^2
mgr(1-cos@)+mglsin@=1/2mv^2
at the top n+mg=mv^2/r
solving the above equations we get the force acting as
6mg(1-cos@+l/rsin@)
c)
let the particle make an angle@ when it leaves the sphere
mgcos@=mv^2/r
mgr(1-cos@)=1/2mv^2
solving it we get cos inverse 2/3
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6 Apr 2008 12:42:11 IST
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51)
let the block A start loosing contact with the surface below it at A' after traveling a distance x( fig 1)
in this process the block b will shift from B to B' such that BB' =AA'=x ( as string is inextensible). and hence there is a loss of gravitational potential energy=mgx
This energy is stored in the spring which is stretched by
L and partly appears as kinetic energy of block A and block B. so by conservation of energy we have
mgx=1/2mv^{2} + 1/2mv^2+ 1/2 k(
L)^2
this gives v^2=gx- k/2m(L)^2- --------------------1
now for vertical equilibrium of A at A'
N+Fcos
=mg
but F=k(L)
and N=0
so k(L)cos=mg---------------------------------------------------------2
further by geometry we get
(L)=L/cos-L=L[1/cos-1]--------------------------3
so substituting the value of (L) from eqn 3 in 2 and solving for cos, we get
cos=4/5
(L)=L/cos-L = 0.1 m
by geometry x=L tan
x=0.3 m
substituting the value of (L) and x in eqn 1 get
V=1.5 m/s
FIG 1
let the block A start loosing contact with the surface below it at A' after traveling a distance x( fig 1)
in this process the block b will shift from B to B' such that BB' =AA'=x ( as string is inextensible). and hence there is a loss of gravitational potential energy=mgx
This energy is stored in the spring which is stretched by
L and partly appears as kinetic energy of block A and block B. so by conservation of energy we have mgx=1/2mv^{2} + 1/2mv^2+ 1/2 k(
L)^2this gives v^2=gx- k/2m(
L)^2- --------------------1now for vertical equilibrium of A at A'
N+Fcos
=mgbut F=k(
L)and N=0
so k(
L)cos=mg---------------------------------------------------------2further by geometry we get
(
L)=L/cos-L=L[1/cos-1]--------------------------3so substituting the value of (
L) from eqn 3 in 2 and solving for cos, we get cos
=4/5(
L)=L/cos-L = 0.1 mby geometry x=L tan
x=0.3 m
substituting the value of (
L) and x in eqn 1 getV=1.5 m/s
FIG 1
11 Apr 2008 16:35:33 IST
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SUM N0 58 c)
IF THE BOB MOVES U]IN A CENTERE WITH THE BO THEN THR RADIUS WILL BE L-X
AT THE HIGHEST POINT
MV^2/R=MG
V^2=G(L-X)
APPLYING THE ENERGY PRINCIPLES
AT THE HIGHEST POINT
1/2MV^2+2MG(L-X)=INITIAL ENERGY
INITIA ENERGY=MGL
SOVR IT AND GET THE ANSWER
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length of the spring at the given instant is h/cos37 = 5h/4
That means the elongation is h/4.
(1/2)kh2/16 = (1/2)mv2
You get the answer to be v=(h/4)(sqrt(k/m))
~Cheerio!!!