A body initially at rest and sliding along a frictionless track from a h...
Neet 2018 | W P E Q 2 | In case of vertical circular motion , we require a
minimum velocity at lowest position of circle to complete the circle. For any
value of velocity greater than this , particle can easily complete its circular
motion and at the minimum value of this velocity , normal force ( in other
cases it could be tension force ) becomes zero .Even at zero normal reaction
force ( in other cases it could be tension force ) particle will not fall
towards the centre since it will has some velocity So it will cross that weightlessness . you must
know about this if you had subscribed my channel because I already posted these
conditions in my twisted qus videos with solution of Q 3 Aipmt/Neet 2004 , year
2008 and year 2016 also.
In this case Mechanical
energy will remain conserved because normal will always act perpendicular to
displacement i.e direction of velocity so its work done will be zero. Gravity
is a conservative force and there is no non-conservative force present .
Conservation of Mechanical Energy
The work-energy theorem, shows that the change DKE in a
body’s kinetic energy is equal to the net work done on it :
DKE = Wnet
Consider separately the work Wc done by conservative force and the work Wnc done
by nonconservative forces. Then
DKE = Wc+ Wnc
We’ve defined the change in potential energy DU as the
negative of the work done by conservative forces. So we can write
DKE = – DU + Wnc
or DKE + DU = Wnc
We define the sum of the kinetic and potential energy as the mechanical energy.
Then Equation shownthat the change in mechanical energy is equal to the work
downe by non-conservative forces.
i.e. DE = Wnc
D \E = 0 if Wnc= 0
Thus if work done by non-conservative forces is zero the mechanical energy of
the system is unchanged.
This called law of conservation of mechanical energy. It may also be written as
DU + DK E = 0
or DU = – DKE
or U + K E = constant
or Uin + K Ein = Uf + K Ef
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