Two very small balls a and b of mass 4 kg and 5 kg are affected to a e...
Two very small balls a and b of mass 4 kg and 5 kg are affected to a e...
Introduction:
We have two small balls, A and B, connected by a light inextensible chord. Ball A has a mass of 4 kg, while ball B has a mass of 5 kg. The chord passes through a frictionless ring fixed at some height above the ground. Ball A is pulled aside and given a horizontal velocity, causing it to move on a circular path parallel to the ground. We need to find the speed of the ball in this equilibrium state.
Analysis:
To find the speed of the ball, we need to consider the forces acting on each ball and analyze the equilibrium condition.
Forces acting on ball A:
1. Tension in the chord: The tension in the chord acts towards the center of the circular path and provides the necessary centripetal force for ball A to move in a circular motion.
2. Weight of ball A: The weight acts vertically downward and can be ignored in this case since the chord is parallel to the ground.
Forces acting on ball B:
1. Tension in the chord: The tension in the chord acts towards the center of the circular path and provides the necessary centripetal force for ball B to remain in equilibrium.
2. Weight of ball B: The weight acts vertically downward and can be ignored in this case since the chord is parallel to the ground.
Equilibrium condition:
For ball B to remain in equilibrium, the net force acting on it must be zero. Therefore, the tension in the chord must be equal to the weight of ball B.
Solution:
Since the tension in the chord is the same for both balls, we can equate the tension in the chord to the weight of ball B:
Tension = Weight of ball B
T = mB * g
Substituting the values, we get:
T = (5 kg) * (9.8 m/s^2) = 49 N
The tension in the chord provides the necessary centripetal force for both balls. Therefore, the speed of the ball is given by:
Tension = (mB * v^2) / r
Substituting the values, we can solve for the speed:
49 N = (5 kg) * v^2 / r
As the radius of the ring is negligible compared to the length of the chord, we can assume r to be approximately infinity. Therefore, the speed of the ball is:
v = sqrt((49 N * r) / (5 kg))
Since the radius is negligible, the speed of the ball will be very close to:
v ≈ sqrt((49 N * infinity) / (5 kg))
v ≈ ∞
Hence, the speed of the ball is closest to infinity.
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