A particle is moving in a circle of radius R in such a way that at any...
Analysis:
To solve this problem, we can use the concepts of centripetal acceleration and tangential acceleration in circular motion. Let's break down the problem step by step.
Given:
- Particle is moving in a circle of radius R.
- At any instant, the normal and tangential component of its acceleration are equal.
- Speed at t = 0 is v0.
Centripetal Acceleration:
The centripetal acceleration is the acceleration directed towards the center of the circle. It is given by the equation:
ac = v^2 / R
Tangential Acceleration:
The tangential acceleration is the acceleration along the tangent to the circle. It is given by the equation:
at = dv / dt
Normal Component of Acceleration:
The normal component of acceleration is the component of acceleration perpendicular to the velocity vector. It is given by the equation:
an = v^2 / R
Equal Components of Acceleration:
According to the given condition, the normal and tangential components of acceleration are equal. Therefore, we can equate the two equations:
v^2 / R = dv / dt
Differential Equation and Time:
Rearranging the equation, we get:
v dv = R dt
Integrating both sides, we have:
∫v dv = ∫R dt
Integrating, we get:
(v^2 / 2) = Rt + C
At t = 0, v = vo, so substituting these values, we get:
(vo^2 / 2) = C
Therefore, the equation becomes:
(v^2 / 2) = Rt + (vo^2 / 2)
Simplifying, we get:
v^2 = 2Rt + vo^2
Time taken for First Revolution:
To find the time taken for the first revolution, we need to find the time when the particle returns to its initial position.
When the particle completes one revolution, it means it has traveled a distance of 2πR. At this point, the displacement is zero. Therefore, we can use the equation of motion:
s = ut + (1/2)at^2
Substituting the values:
2πR = v0t + (1/2)at^2
Since the particle is moving in a circle, the displacement is equal to the circumference of the circle, which is 2πR. The initial velocity, u, is v0, and the acceleration, a, is v^2 / R.
Substituting these values, we get:
2πR = v0t + (1/2)((v^2 / R)t^2)
Simplifying, we get:
2πR = v0t + (1/2)(2Rt + vo^2)t
2πR = v0t + Rt^2 + (vo^2 / 2)t
This is a quadratic equation in t. Solving it will give us the time taken to complete the first revolution.
Conclusion:
In this problem, we used the concepts of centripetal acceleration and tangential acceleration to find the time taken to complete the first revolution. We started by equating the normal and tangential components of acceleration and obtained
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.