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A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b?
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A particle moves in a circular path of radius R with an angular veloci...
Acceleration of the Particle in a Circular Path
- Given Parameters:
- Radius of the circular path, R
- Angular velocity, ω = a - bt
- Acceleration of the Particle:
- The acceleration of the particle moving in a circular path is given by the centripetal acceleration, which is directed towards the center of the circle.
- The centripetal acceleration (a_c) is given by the formula: a_c = Rω^2
- Calculating Acceleration at time 2a/b:
- To find the magnitude of the acceleration of the particle after time 2a/b, we need to substitute t = 2a/b into the given angular velocity function.
- ω = a - b(2a/b) = a - 2a = -a
- Therefore, the angular velocity at time 2a/b is -a.
- Substitute into Centripetal Acceleration Formula:
- Substitute ω = -a into the centripetal acceleration formula: a_c = R(-a)^2 = Ra^2
- Magnitude of Acceleration:
- The magnitude of acceleration is the absolute value of the centripetal acceleration, so the magnitude of acceleration at time 2a/b is |Ra^2| = Ra^2
- Conclusion:
- The magnitude of the acceleration of the particle at time 2a/b is Ra^2 in the circular path of radius R with angular velocity ω = a - bt.
- Given Parameters:
- Radius of the circular path, R
- Angular velocity, ω = a - bt
- Acceleration of the Particle:
- The acceleration of the particle moving in a circular path is given by the centripetal acceleration, which is directed towards the center of the circle.
- The centripetal acceleration (a_c) is given by the formula: a_c = Rω^2
- Calculating Acceleration at time 2a/b:
- To find the magnitude of the acceleration of the particle after time 2a/b, we need to substitute t = 2a/b into the given angular velocity function.
- ω = a - b(2a/b) = a - 2a = -a
- Therefore, the angular velocity at time 2a/b is -a.
- Substitute into Centripetal Acceleration Formula:
- Substitute ω = -a into the centripetal acceleration formula: a_c = R(-a)^2 = Ra^2
- Magnitude of Acceleration:
- The magnitude of acceleration is the absolute value of the centripetal acceleration, so the magnitude of acceleration at time 2a/b is |Ra^2| = Ra^2
- Conclusion:
- The magnitude of the acceleration of the particle at time 2a/b is Ra^2 in the circular path of radius R with angular velocity ω = a - bt.
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A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b?.
A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a circular path of radius R with an angular velocity w=a-bt. The magnitude of the acceleration of the particle after time 2a/b?.
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