Tangential acceleration of a particle moving in circle of radius 1m va...
Tangential Acceleration and Radial Acceleration
In order to solve this problem, we first need to understand the concepts of tangential acceleration and radial acceleration.
Tangential acceleration refers to the rate of change of the magnitude of the velocity vector of a particle moving in a circular path. It is always directed tangent to the path and can be calculated using the formula:
a_t = d(v)/dt
where a_t is the tangential acceleration, v is the velocity, and t is time.
Radial acceleration refers to the acceleration of a particle along the radial direction of the circular path. It is always directed towards the center of the circle and can be calculated using the formula:
a_r = v^2/r
where a_r is the radial acceleration, v is the velocity, and r is the radius of the circle.
Finding the Total Acceleration
To find the total acceleration, we can use the Pythagorean theorem as the tangential acceleration and radial acceleration are perpendicular to each other. The total acceleration (a) can be calculated using the formula:
a = sqrt(a_t^2 + a_r^2)
where a is the total acceleration, a_t is the tangential acceleration, and a_r is the radial acceleration.
Angle between Total Acceleration and Radial Acceleration
Let's assume that the tangential acceleration of the particle varies with time t as given in the problem. We need to find the time after which the total acceleration makes an angle of 30 degrees with the radial acceleration.
We can represent the angle between the total acceleration and the radial acceleration using the equation:
tan(theta) = a_t / a_r
where theta is the angle between the total acceleration and the radial acceleration.
To find the time, we need to solve this equation for t. Rearranging the equation, we have:
tan(theta) = a_t / (v^2/r)
Substituting the expression for tangential acceleration (a_t) and radial acceleration (a_r) into the equation, we get:
tan(theta) = (d(v)/dt) / (v^2/r)
Simplifying the equation, we have:
tan(theta) = (r * dv) / (v^2)
Integrating both sides of the equation with respect to v and applying the limits, we can solve for t:
∫ (1/(r * tan(theta))) dv = ∫ (1/v^2) dv
Integrating both sides, we get:
(1/(r * tan(theta))) * ln(v) = -1/v + C
where C is the constant of integration.
Solving for v, we have:
v = e^(-(1/(r * tan(theta))) * ln(v) + C)
Simplifying the equation, we get:
v = e^(C - (1/(r * tan(theta))) * ln(v))
Let's assume the initial velocity of the particle is v_0. Substituting this value into the equation, we have:
v_0 = e^(C - (1/(r * tan(theta))) * ln(v_0))
Taking the natural logarithm of both sides and solving for C, we get:
C = ln(v_0) + (1/(r * tan(theta))) * ln(v_0)
Substituting the value of C
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.