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A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :
- a)angular velocity of rod is increased continuously
- b)ω = 10 rad/s
- c)angular velocity of rod is decreased continuously
- d)constant velocity of ring is not possible
Correct answer is option 'B'. Can you explain this answer?
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A ring attached with a light spring is fitted in a smooth rod. The spr...
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A ring attached with a light spring is fitted in a smooth rod. The spr...
And the spring constant is 10 N/m. The rod is initially horizontal and the spring is unstretched.
To find the tension in the spring when the rod is in a vertical position, we need to consider the forces acting on the ring.
When the rod is horizontal, the only force acting on the ring is its weight, which is given by the equation F = mg, where m is the mass of the ring (3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of the ring is F = 3 kg × 9.8 m/s^2 = 29.4 N.
When the rod is vertical, the tension in the spring counteracts the weight of the ring. The tension in the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the equilibrium position of the spring is when it is unstretched.
The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant (10 N/m), and x is the displacement from the equilibrium position. Since the spring is attached at the outer end of the rod, the displacement x is equal to the length of the rod.
The length of the rod can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the length of the rod, and the other two sides are the height and base of the triangle formed by the rod.
Let's assume the height of the triangle is h and the base is b. Since the rod is initially horizontal, the height is equal to the length of the rod when it is vertical. Therefore, h = b.
Using the Pythagorean theorem, we can write the equation h^2 + b^2 = L^2, where L is the length of the rod.
Since h = b, we can simplify the equation to 2h^2 = L^2, or h = sqrt(L^2 / 2).
Since the rod is initially horizontal, the displacement x when it is vertical is equal to the length of the rod, which is h = sqrt(L^2 / 2).
Substituting the values into Hooke's Law, we get F = -kx = -10 N/m × sqrt(L^2 / 2).
Since the tension in the spring counteracts the weight of the ring, the tension in the spring when the rod is vertical is equal to the weight of the ring plus the force exerted by the spring, which is given by F = 29.4 N - (-10 N/m × sqrt(L^2 / 2)).
Therefore, the tension in the spring when the rod is vertical is 29.4 N + 10 N/m × sqrt(L^2 / 2).
To find the tension in the spring when the rod is in a vertical position, we need to consider the forces acting on the ring.
When the rod is horizontal, the only force acting on the ring is its weight, which is given by the equation F = mg, where m is the mass of the ring (3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of the ring is F = 3 kg × 9.8 m/s^2 = 29.4 N.
When the rod is vertical, the tension in the spring counteracts the weight of the ring. The tension in the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the equilibrium position of the spring is when it is unstretched.
The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant (10 N/m), and x is the displacement from the equilibrium position. Since the spring is attached at the outer end of the rod, the displacement x is equal to the length of the rod.
The length of the rod can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the length of the rod, and the other two sides are the height and base of the triangle formed by the rod.
Let's assume the height of the triangle is h and the base is b. Since the rod is initially horizontal, the height is equal to the length of the rod when it is vertical. Therefore, h = b.
Using the Pythagorean theorem, we can write the equation h^2 + b^2 = L^2, where L is the length of the rod.
Since h = b, we can simplify the equation to 2h^2 = L^2, or h = sqrt(L^2 / 2).
Since the rod is initially horizontal, the displacement x when it is vertical is equal to the length of the rod, which is h = sqrt(L^2 / 2).
Substituting the values into Hooke's Law, we get F = -kx = -10 N/m × sqrt(L^2 / 2).
Since the tension in the spring counteracts the weight of the ring, the tension in the spring when the rod is vertical is equal to the weight of the ring plus the force exerted by the spring, which is given by F = 29.4 N - (-10 N/m × sqrt(L^2 / 2)).
Therefore, the tension in the spring when the rod is vertical is 29.4 N + 10 N/m × sqrt(L^2 / 2).
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A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? for LR 2025 is part of LR preparation. The Question and answers have been prepared according to the LR exam syllabus. Information about A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for LR 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer?.
A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? for LR 2025 is part of LR preparation. The Question and answers have been prepared according to the LR exam syllabus. Information about A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for LR 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for LR.
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Here you can find the meaning of A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A ring attached with a light spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is 3kg & spring constant of spring is 300 N/m. The ring is given a velocity ‘V’ towards the outer end of the rod and the rod is set to be rotating with an angular velocity ω. Then ring will move with constant speed with respect to the rod if :a)angular velocity of rod is increased continuouslyb)ω = 10 rad/s c)angular velocity of rod is decreased continuouslyd)constant velocity of ring is not possibleCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice LR tests.
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