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Work Power Energy -5
Derivation:
Motion in a vertical circle:
A particle of mass m is attached to inextensible light string of length l and it is imparted a velocity u in horizontal direction lowest point. Let v be its velocity at point B of circle.
Fig (23)
Here h = R (1-Cosq) ------------ (1)
By conservation of mechanical energy
Necessary centripetal force is provided by the resultant of tension T and mgCos
Now three condition arise depending on u,
The string does not stuck at highest point if T 0 at =,
So substituting T = 0 and = for completing the circle in (B)
(At highest point)
And h = 2R
So from equation (2)
Or
Or
the particle will complete the circle.
Substituting q = 00 and v = u = in equation (3) we get T = 6mg. So in the critical condition tension in the string at lowest position is 6mg.
Fig (24)
Putting T = 0 before reaching highest point for
Substituting in equation (1) we get
Velocity of particle becomes zero when
0 = u2-2gh
Now the particle will leave the circle if tension in the string becomes zero but velocity is not zero or,
T = 0 but v 0 this is possible only when
h1<h2
Therefore if the particle leaves the circle.
The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.
V = 0 but T 0 this is possible only when
h2 < h1
Further if h1 = h2, and tension and velocity both becomes zero simultaneously.
Illustration:
A ball of mass m slides without friction down a path from height h and then moves in loop of radius Find the force exerted on the ball by the track at B and at C
Fig (27)
Solution:
Fig (28)
For maintaining circular motion net centripetal force must be greater than
On applying energy conservation between point A and C we get
On solving
From f.b.d of body
On solving FB=3mg (1+Cosb)
Similarly for point B
mg [h-(R+RCosa)] = (energy equilibrium)
FB = 3mg (1-Cosa)