Tangential acceleration of a particle moving in a circle of radius 1m ...
Problem Statement:
The tangential acceleration of a particle moving in a circle of radius 1m varies with time t, and the initial velocity of the particle is zero. We need to determine the time after which the total acceleration of the particle makes an angle of 30 degrees with the radial acceleration.
Solution:
To solve this problem, we need to understand the components of acceleration in circular motion and the relationship between them.
Components of Acceleration:
In circular motion, the acceleration of a particle can be divided into two components: radial acceleration and tangential acceleration.
1. Radial Acceleration (ar): It is the acceleration that acts towards the center of the circle and is responsible for changing the direction of velocity. The magnitude of radial acceleration is given by ar = v^2 / r, where v is the magnitude of velocity and r is the radius of the circle.
2. Tangential Acceleration (at): It is the acceleration that acts tangentially to the circle and is responsible for changing the magnitude of velocity. The magnitude of tangential acceleration is given by at = dv / dt, where dv is the change in velocity and dt is the change in time.
Relationship between Tangential and Radial Acceleration:
The total acceleration (a) of the particle can be found using the Pythagorean theorem as follows:
a = √(ar^2 + at^2)
The angle (θ) between the total acceleration and the radial acceleration can be found using the formula:
θ = tan^(-1)(at / ar)
Finding the Time:
We are given that the tangential acceleration varies with time t. Let's assume the tangential acceleration as at = kt, where k is a constant.
Substituting the value of tangential acceleration in the formula of total acceleration, we get:
a = √(ar^2 + (kt)^2)
The angle θ between the total acceleration and the radial acceleration is given as 30 degrees. Converting it to radians, we have θ = 30 * π / 180 = π / 6 radians.
Using the formula of the tangent of an angle, we can write:
tan(π / 6) = (kt) / ar
Simplifying the equation, we get:
√3 = (kt) / ar
Substituting the value of radial acceleration as ar = v^2 / r, we have:
√3 = (kt) / (v^2 / r)
Simplifying further, we get:
v = √(kt / (√3 * r))
Since the initial velocity of the particle is zero, we can set v = 0 and solve for t as follows:
0 = √(kt / (√3 * r))
Squaring both sides, we get:
0 = kt / (√3 * r)
Simplifying further, we have:
kt = 0
Therefore, the time after which the total acceleration of the particle makes an angle of 30 degrees with the radial acceleration is t = 0.
Tangential acceleration of a particle moving in a circle of radius 1m ...
Tangential/radial=√3
a/w^2=√3
√3*dw/dt=w^2
now u can solve for time
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