Derive an expression for linear acceleration of a partical performing UCM
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Answer:
Derive an expression for acceleration of particle performing uniform circular motion?
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.now, as.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.now, as.F = ma.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.now, as.F = ma.thus,
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.now, as.F = ma.thus,ma = mv^2/r. so, linear acceleration will be.
Derive an expression for acceleration of particle performing uniform circular motion?If an object moves in a circle it changes its velocity i.e. it is accelerating.we know that centripetal force is given as.F = mv^2/r.now, as.F = ma.thus,ma = mv^2/r. so, linear acceleration will be.a = v^2/r = w^2r.
Explanation:
ircular path is called as centripetal acceleration. The centripetal acceleration is directed along the radius and is also called radial acceleration. Expression for acceleration in U.C.M by analytical method (Geometrical method): i. Consider a particle performing uniform circular motion in a circle of centre O and radius r with a uniform linear velocity of magnitude v. ii. Let a particle travel a very short distance δs from A to B in a very short time interval δt. iii. Let δθ be the angle described by the radius vector OA in the time interval δt as shown in the figure. iv. The velocities at A and B are directed along the tangent. v. Velocity at B is represented by ¯¯¯¯¯¯¯¯ B C BC¯ while the velocity at A is represented by ¯¯¯¯¯¯¯¯¯¯ A M AM¯. [Assuming AM = BD] vi. Angle between ¯¯¯¯¯¯¯¯ B C BC¯ and ¯¯¯¯¯¯¯¯¯ B D BD¯ is equal to δθ as they are perpendicular to ¯¯¯¯¯¯¯¯ O B OB¯ and ¯¯¯¯¯¯¯¯ O A OA¯ respectively. vii. Since ∆BDC ~OAB viii. For very small δt, arc length δs of circular path between A and B can be taken as AB As δt → 0, B approaches A and δv becomes perpendicular to the tangent i.e., along the radius towards the centre. x. In vector form, → a = − ω 2 → r a→=−ω2r→ Negative sign shows that direction of → a a→ is opposite to the direction of → r r→. Read more on Sarthaks.com - https://www.sarthaks.com/1201355/derive-an-expression-for-linear-acceleration-of-a-particle-performing-u-c-m?show=1201357