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Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to?
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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.
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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Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to?
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Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to?.
Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very small balls a and b of mass 4 kg and 5 kg are affected to a ends of a light inextensible chord that passes through a frictionless ring of radius negligible compared to the length of the chord that is fixed at some height above the ground ball a is pulled aside and given horizontal velocity so that it is starts moving on a circular path parallel to the ground keeping bol be in equilibrium is speed of the ball is closest to?.
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