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Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to?
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Two very small balls a and b of masses 4 kg and 5 kg are a fixed to th...
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Two very small balls a and b of masses 4 kg and 5 kg are a fixed to th...
Problem Analysis:
We have two small balls, A and B, of masses 4 kg and 5 kg respectively. These balls are attached to the ends of a light inextensible cord, which passes through a frictionless ring. The ring is fixed at some height above the ground. When the cord is pulled aside and given a horizontal velocity, it starts moving on a circular path parallel to the ground, while keeping ball A at the center.
Solution:
Step 1: Analyzing the Forces:
When the cord is pulled aside and given a horizontal velocity, the system will experience various forces:
1. Tension in the cord: The tension in the cord will provide the necessary centripetal force to keep ball B moving in a circular path. This tension force can be calculated using the equation: T = mv^2 / r, where m is the mass of ball B, v is its velocity, and r is the radius of the circular path.
2. Weight of ball B: The weight of ball B acts vertically downwards, and its magnitude can be calculated as W = mg, where m is the mass of ball B and g is the acceleration due to gravity.
3. Tension in the cord at the center: Since ball A is in equilibrium at the center, the tension in the cord at the center must be equal to the weight of ball B.
Step 2: Calculating the Tension in the Cord:
Since ball A is in equilibrium at the center, the tension in the cord at the center must be equal to the weight of ball B. Therefore, we can equate the tension in the cord at the center with the weight of ball B:
T = mg
Substituting the values, we get:
T = 5 kg * 9.8 m/s^2 = 49 N
Step 3: Calculating the Velocity of Ball A:
Since ball A is at the center and in equilibrium, the tension in the cord at the center provides the necessary centripetal force to keep ball A in place. Therefore, the tension in the cord at the center is equal to the centripetal force required for ball A:
T = mv^2 / r
Substituting the values, we get:
49 N = 4 kg * v^2 / r
Simplifying the equation, we get:
v^2 = (49 N * r) / 4 kg
Taking the square root of both sides, we get:
v = sqrt((49 N * r) / 4 kg)
Step 4: Finding the Closest Value of the Speed:
To find the closest value of the speed of ball A, we need to know the value of the radius of the circular path. Unfortunately, the problem does not provide this information. Therefore, we cannot determine the exact value of the speed of ball A. We can only calculate the speed if we know the radius of the circular path.
Conclusion:
In this problem, we analyzed the forces acting on the system of two small balls attached to a cord passing through a frictionless ring. We calculated the tension in the cord at the center and used it to find the velocity of ball A. However, without knowing the radius of the circular path, we cannot determine the exact value of the speed of ball A.
We have two small balls, A and B, of masses 4 kg and 5 kg respectively. These balls are attached to the ends of a light inextensible cord, which passes through a frictionless ring. The ring is fixed at some height above the ground. When the cord is pulled aside and given a horizontal velocity, it starts moving on a circular path parallel to the ground, while keeping ball A at the center.
Solution:
Step 1: Analyzing the Forces:
When the cord is pulled aside and given a horizontal velocity, the system will experience various forces:
1. Tension in the cord: The tension in the cord will provide the necessary centripetal force to keep ball B moving in a circular path. This tension force can be calculated using the equation: T = mv^2 / r, where m is the mass of ball B, v is its velocity, and r is the radius of the circular path.
2. Weight of ball B: The weight of ball B acts vertically downwards, and its magnitude can be calculated as W = mg, where m is the mass of ball B and g is the acceleration due to gravity.
3. Tension in the cord at the center: Since ball A is in equilibrium at the center, the tension in the cord at the center must be equal to the weight of ball B.
Step 2: Calculating the Tension in the Cord:
Since ball A is in equilibrium at the center, the tension in the cord at the center must be equal to the weight of ball B. Therefore, we can equate the tension in the cord at the center with the weight of ball B:
T = mg
Substituting the values, we get:
T = 5 kg * 9.8 m/s^2 = 49 N
Step 3: Calculating the Velocity of Ball A:
Since ball A is at the center and in equilibrium, the tension in the cord at the center provides the necessary centripetal force to keep ball A in place. Therefore, the tension in the cord at the center is equal to the centripetal force required for ball A:
T = mv^2 / r
Substituting the values, we get:
49 N = 4 kg * v^2 / r
Simplifying the equation, we get:
v^2 = (49 N * r) / 4 kg
Taking the square root of both sides, we get:
v = sqrt((49 N * r) / 4 kg)
Step 4: Finding the Closest Value of the Speed:
To find the closest value of the speed of ball A, we need to know the value of the radius of the circular path. Unfortunately, the problem does not provide this information. Therefore, we cannot determine the exact value of the speed of ball A. We can only calculate the speed if we know the radius of the circular path.
Conclusion:
In this problem, we analyzed the forces acting on the system of two small balls attached to a cord passing through a frictionless ring. We calculated the tension in the cord at the center and used it to find the velocity of ball A. However, without knowing the radius of the circular path, we cannot determine the exact value of the speed of ball A.
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Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to?
Question Description
Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to?.
Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two very small balls a and b of masses 4 kg and 5 kg are a fixed to the ends of a light inextensible cord that passes through a frictionless ring of radius negligible compared to the length of the cord. the ring is fixed at some height above the ground. it is pulled aside and given a horizontal velocity so that it start moving on a circular path parallel to the Ground. keeping ball a in equilibrium at centre. the speed of the ball a is closest to?.
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