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        Derivation of  centripetal acceleration?
7 months ago

Arun
13149 Points
							 Motion of a Particle in a Circular PathIt is a special kind of two-dimensional motion in which the particle's position vector always lies on the circumference of a circle. In order to calculate the acceleration parameter it is helpful to first consider circular motion with constant speed, calleduniform circular motion. Let there be a particle moving along a circle of radius r with a velocity , as shown in figure given below, such that  = v = constant. For this particle, it is our aim to calculate the magnitude and direction of its acceleration. We know that,Now, we have to find an expression for  in terms of known quantities. For this, consider the particle velocity vector at two points A and B. Displacing , parallel to itself and placing it back to back with , as shown in figure given below. We have =  – Consider ΔAOB, angle between OA and OB is same as angle between  and  because  is perpendicular to vector OA and  is also perpendicular to vector OB.OB = OA = r andSo, ΔAOB is similar to the triangle formed by , and Thus,From geometry we have, Δv/v = AB/rNow AB is approximately equal to vΔt.In the limit Δt → 0 the above relation becomes exact, we haveThis is the magnitude of the acceleration. The direction  is instantaneously along a radius inward towards the centre of the circle, because of this  is called radial or centripetal acceleration. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

7 months ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions