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Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:?
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Two particles start moving on the same circle of radius 2m , from the ...
Introduction:
In this problem, we have two particles moving on the same circle with different tangential accelerations. We need to find the point where they meet for the first time and calculate the smaller angle subtended by the line joining the meeting point and the center of the circle.
Given:
- Radius of the circle = 2m
- Tangential acceleration of Particle 1 (clockwise) = 2m/s^2
- Tangential acceleration of Particle 2 (anticlockwise) = 6m/s^2
Approach:
To solve this problem, we will first calculate the linear speed of each particle using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Then, we will find the time taken by each particle to reach the point of meeting on the circle using the formula s = ut + (1/2)at^2, where s is the distance covered.
Next, we will calculate the angle subtended by each particle at the center of the circle using the formula θ = s/r, where θ is the angle, s is the arc length, and r is the radius of the circle.
Finally, we will compare the angles subtended by each particle and determine the smaller angle.
Calculations:
1. Particle 1 (clockwise):
- Initial velocity (u) = 0 m/s (as it starts from rest)
- Final velocity (v) = u + at = 0 + 2(1) = 2 m/s
- Time taken (t) = (v - u) / a = (2 - 0) / 2 = 1 s
- Distance covered (s) = ut + (1/2)at^2 = 0(1) + (1/2)(2)(1)^2 = 1 m
- Angle subtended (θ1) = s / r = 1 / 2 = 0.5 radians
2. Particle 2 (anticlockwise):
- Initial velocity (u) = 0 m/s (as it starts from rest)
- Final velocity (v) = u + at = 0 + 6(1) = 6 m/s
- Time taken (t) = (v - u) / a = (6 - 0) / 6 = 1 s
- Distance covered (s) = ut + (1/2)at^2 = 0(1) + (1/2)(6)(1)^2 = 3 m
- Angle subtended (θ2) = s / r = 3 / 2 = 1.5 radians
Conclusion:
The smaller angle subtended by the line joining the meeting point (Q) and the center of the circle is 0.5 radians.
In this problem, we have two particles moving on the same circle with different tangential accelerations. We need to find the point where they meet for the first time and calculate the smaller angle subtended by the line joining the meeting point and the center of the circle.
Given:
- Radius of the circle = 2m
- Tangential acceleration of Particle 1 (clockwise) = 2m/s^2
- Tangential acceleration of Particle 2 (anticlockwise) = 6m/s^2
Approach:
To solve this problem, we will first calculate the linear speed of each particle using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Then, we will find the time taken by each particle to reach the point of meeting on the circle using the formula s = ut + (1/2)at^2, where s is the distance covered.
Next, we will calculate the angle subtended by each particle at the center of the circle using the formula θ = s/r, where θ is the angle, s is the arc length, and r is the radius of the circle.
Finally, we will compare the angles subtended by each particle and determine the smaller angle.
Calculations:
1. Particle 1 (clockwise):
- Initial velocity (u) = 0 m/s (as it starts from rest)
- Final velocity (v) = u + at = 0 + 2(1) = 2 m/s
- Time taken (t) = (v - u) / a = (2 - 0) / 2 = 1 s
- Distance covered (s) = ut + (1/2)at^2 = 0(1) + (1/2)(2)(1)^2 = 1 m
- Angle subtended (θ1) = s / r = 1 / 2 = 0.5 radians
2. Particle 2 (anticlockwise):
- Initial velocity (u) = 0 m/s (as it starts from rest)
- Final velocity (v) = u + at = 0 + 6(1) = 6 m/s
- Time taken (t) = (v - u) / a = (6 - 0) / 6 = 1 s
- Distance covered (s) = ut + (1/2)at^2 = 0(1) + (1/2)(6)(1)^2 = 3 m
- Angle subtended (θ2) = s / r = 3 / 2 = 1.5 radians
Conclusion:
The smaller angle subtended by the line joining the meeting point (Q) and the center of the circle is 0.5 radians.
Community Answer
Two particles start moving on the same circle of radius 2m , from the ...
Answer should be right angle.
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Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:?
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Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:?.
Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:?.
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Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:?, a detailed solution for Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? has been provided alongside types of Two particles start moving on the same circle of radius 2m , from the same point P at t=0 , with constant tangential acceleration 2m/s^2 and 6m/s^2 , clockwise and anticlockwise, respectively. The point where they meet for the first time is Q. The smaller angle subtended by PQ at center of circle is:? theory, EduRev gives you an
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