A semi circular rail of horizontal diameter AB=2R is placed in a vertical plane. Two particles S1 and S2 of respective masses m1 and m2 (m1 <m2) slide without friction on the inner surface of this rail.
A) Particle S1 is released at point B without initial velocity .Determine its velocity V1 at the lowest point C.
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To determine the velocity V1 of particle S1 at the lowest point C, we can use the principle of conservation of mechanical energy.
The mechanical energy of a particle consists of two components: kinetic energy (KE) and potential energy (PE). Since there is no friction or external forces acting on the particles, the total mechanical energy remains constant throughout the motion.
1. Gravitational Potential Energy:
At point B (highest point), the gravitational potential energy of particle S1 is maximum since it is at the highest vertical position. At the lowest point C, the potential energy is zero. This change in potential energy is given by:
ΔPE = m1 * g * h
Note: g is the acceleration due to gravity and h is the height difference between points B and C.
2. Kinetic Energy:
At point B, the kinetic energy of particle S1 is zero since it is released without initial velocity. At the lowest point C, the kinetic energy is maximum. The kinetic energy of a particle is given by:
KE = 0.5 * m1 * V1^2
Since the total mechanical energy is conserved, we can equate the change in potential energy to the increase in kinetic energy:
ΔPE = KE
m1 * g * h = 0.5 * m1 * V1^2
Simplifying the equation, we can solve for V1:
V1^2 = 2 * g * h
V1 = √(2 * g * h)
Therefore, the velocity V1 of particle S1 at the lowest point C is equal to the square root of twice the acceleration due to gravity (g) multiplied by the height difference (h) between points B and C.
Note: The direction of velocity V1 is downward at point C, as the particle is sliding along the inner surface of the rail and under the influence of gravity.