MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        
Derivation of centripetal acceleration?
4 months ago

Answers : (1)

Arun
11430 Points
							
 

Motion of a Particle in a Circular Path

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 \vec{v}, as shown in figure given below, such that \left | \vec{v} \right | = v = constant. For this particle, it is our aim to calculate the magnitude and direction of its acceleration. We know that,

\vec{a} =\lim_{\Delta t \to 0}\frac{\Delta \vec{v}}{\Delta t}

Now, we have to find an expression for \Delta \vec{v} in terms of known quantities. For this, consider the particle velocity vector at two points A and B. Displacing \vec{v_{B}}, parallel to itself and placing it back to back with \vec{v_{A}}, as shown in figure given below. We have

\Delta \vec{v} = \vec{v_{B}} – \vec{v_{A}}

Consider ΔAOB, angle between OA and OB is same as angle between \vec{v_{A}} and \vec{v_{B}} because \vec{v_{A}} is perpendicular to vector OA and \vec{v_{B}} is also perpendicular to vector OB.

OB = OA = r and

\left | \vec{v_{A}} \right | = \left | \vec{v_{B}} \right | = v

Component of Velocity of the Particle in Circular Path

So, ΔAOB is similar to the triangle formed by \vec{v_{A}},\vec{v_{B}} and \Delta \vec{v}

Thus,From geometry we have, Δv/v = AB/r

Now AB is approximately equal to vΔt.

\frac{\Delta v}{v}\cong \frac{v\Delta t}{r}

\Rightarrow \frac{v^{2}}{r}\cong \frac{\Delta v}{\Delta t}

In the limit Δt → 0 the above relation becomes exact, we have

\left | \vec{a} \right |=\lim_{\Delta t\to0 }\frac{\Delta \vec{v}}{\Delta t}=\frac{v^{2}}{r}

This is the magnitude of the acceleration. The direction \vec{a} is instantaneously along a radius inward towards the centre of the circle, because of this \vec{a} is called radial or centripetal acceleration. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

4 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 551 off

COUPON CODE: SELF20


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

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


Get Extra Rs. 594 off

COUPON CODE: SELF20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details