Conservation of mechanical energy can be applied in cases in which internal forces are conservative and external forces do not work. This also includes the cases where external forces are present but do not work. One of such cases is circular motion in which tension does zero work,
For a bob to just complete the full circle it must satisfy the condition that its velocity at highest point (Z) should follow,
V2/R = g
At this speed, gravity provides necessary centripetal force sufficiently at highest point and tension is no longer required (for that instant).
The speed of the particle required at X necessary for the particle to cross Z with sqrt(Rg) = sqrt(5rg)
The speed of the particle required at Y necessary for the particle to cross Z with sqrt(Rg) = sqrt(3rg)
Q.1 Mark the CORRECT statement(s) from the following options.
(a) If speed of the particle at X is less than or equal to sqrt(2rg) it would oscillate about mean position.
(b) If speed of the particle at X is less than sqrt(5rg) but greater than sqrt(2rg) it would leave the circle after crossing the point Y.
(c) If speed of the particle at X is more than or equal to sqrt(5rg) it would complete the full circle of motion.
(d) If the string in the motion in the vertical circle problem is replaced by a rod, ZERO speed is required for the particle at the highest point (Z) to maintain circular motion.
Q.2 Find v1 and θ, for which tension in the string at the given instant is mg, where ‘m’ is mass of the particle.
Q.3 Find (a) net force on block at Q when released from P (b) At what height above the track (h) should the block be released so that the force on track by the block when it is at R is equal to mg.
Q.4 In the figure drawn below, a horizontal light rod with a hook at one end and a mass ‘m’ attached to the other end is dropped, find minimum ‘h’ so that the block completes the full circle.
Q.5 A particle kept on a smooth hemisphere is released from ‘X’ and normal reaction of the track vanishes at Y then,
(a) sin Φ = cos θ (b) sin Φ = 0.5 cos θ (c) sin Φ = 0.66 cos θ (d) sin Φ = 0.75 cos θ
Q.1 The potential energy U (in joule) of a particle of mass 1 kg moving in x-y plane obeys the law U = 3x + 4y, x & y are in metre. If the particle is at rest at (6,4) when the motion begins then, (a) Find the acceleration of the particle (b) Speed of the particle when it crosses y-axis. (c) Coordinates of particle at t = 1s.
Q.2 The potential energy of a particle is U = k(x^2 + y^2), where k > 0. The particle is at rest at (3,3) & begins to move towards P (1,1). Find its K.E. when it reaches (1,1). (a) 8k (b) 24k (c) 16k (d) zero
Q.3 F(x) = -kx + ax^3, when k and a are positive constants. The particle can move only along x-axis due to F. Plot the U(x) curve given that U(0) = 0.