A particle starts from rest at time t = 20 and move on a circular path...
Problem: A particle starts from rest at time t = 20 and moves on a circular path of radius 1 metre with tangential acceleration root 3 metre per second square. After 1 second, its acceleration makes an angle theta with its velocity. What is the value of 3 tan square theta?
Solution:
Step 1: Find the magnitude of the centripetal acceleration of the particle.
Given, radius of the circular path, r = 1 metre
tangential acceleration, at = root 3 metre per second square
Centripetal acceleration, ac = v^2 / r
where v is the velocity of the particle
Initially, the particle is at rest, so its initial velocity, u = 0
After 1 second, the particle's velocity, v = u + at*t = 0 + root 3 * 1 = root 3 m/s
Therefore, ac = (root 3)^2 / 1 = 3 m/s^2
Step 2: Find the direction of the acceleration and the angle it makes with the velocity.
The acceleration of the particle has two components - tangential and centripetal.
The direction of the centripetal acceleration is towards the centre of the circular path.
The direction of the tangential acceleration is perpendicular to the centripetal acceleration and in the direction of motion of the particle.
So, the direction of the net acceleration of the particle is at an angle theta with its velocity.
Using trigonometry, we can find the value of theta.
tan theta = ac / at = 3 / root 3 = root 3
Therefore, theta = tan inverse (root 3) = 60 degrees
Step 3: Find the value of 3 tan square theta.
3 tan square theta = 3 * tan^2 (60 degrees) = 3 * (root 3)^2 = 9
Therefore, 3 tan square theta = 9.
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