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Uniform circular motion

A particle moves along a circle with a velocity v=k t, where k

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A particle moves along a circle with a velocity $v = k t$ , where $k$ is a constant. Find the net acceleration of the particle at the instant when it has covered $n$ th fraction of the circle after the beginning of motion.

Name: Video solution 1: A particle moves along a circle with a velocity v=kt, where k is a constant. Find the net acceleration of the particle at the instant when it has covered nth fraction of the circle after the beginning of motion.
Uploaded: 2022-08-30T04:39:38.000Z
Duration: 4 min
Description: Video solution 1: A particle moves along a circle with a velocity v=kt, where k is a constant. Find the net acceleration of the particle at the instant when it has covered nth fraction of the circle after the beginning of motion.

Views: 5,327 students

Updated on: Sep 4, 2023

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IIT JEE Advanced Comprehensive Physics (McGraw Hill)

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Levi Discussed

A particle moves along a circle with a velocity

v = k t

, where

k

is a constant. Find the net acceleration of the particle at the instant when it has covered

n

th fraction of the circle after the beginning of motion.

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Let

v

be the speed of the particle at the instant when it has covered

n

th fraction of the circle of radius

R

. Centripetal acceleration at that instant is

a_{c} = \frac{v ^{2}}{R}

(1) Tangential acceleration at that instant is

a_{t} = \frac{d v}{d t} = \frac{d v}{d t} \frac{d x}{d x} = \frac{v d v}{d x}

\Rightarrow v d v = a_{t} d x

(2) Let

x

be the distance covered. Then

x = (2 π R) n

. Integrating Eq. (2) we have

\int_{0}^{v} v d v = a_{t} \int_{0}^{x} d x

\frac{v ^{2}}{2} = a_{t} x = a_{t} (2 π R) n

v = 2 a_{t} (2 π R) n

(3) Using (3) in (1), we get

a_{c} = \frac{2 a _{t} ( 2 π R ) n}{R} = 4 π a_{t} n

Now

a_{t} = \frac{d v}{d t} = \frac{d}{d t} (k t) = k

a_{c} = 4 πkn

∴

Net acceleration

a = a_{t}^{2} + a_{c}^{2} = k^{2} + (4 πkn)^{2} = k 1 + 16 π^{2} n^{2}

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Question Text	A particle moves along a circle with a velocity $v = k t$ , where $k$ is a constant. Find the net acceleration of the particle at the instant when it has covered $n$ th fraction of the circle after the beginning of motion.
Updated On	Sep 4, 2023
Topic	Motion in a Plane
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	277
Avg. Video Duration	6 min

A particle moves along a circle with a velocity $v = k t$ , where $k$ is a constant. Find the net acceleration of the particle at the instant when it has covered $n$ th fraction of the circle after the beginning of motion.

Text solutionVerified

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A particle moves along a circle with a velocity v=kt, where k is a constant. Find the net acceleration of the particle at the instant when it has covered nth fraction of the circle after the beginning of motion.

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Practice questions from IIT JEE Advanced Comprehensive Physics (McGraw Hill)

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A particle moves along a circle with a velocity $v = k t$ , where $k$ is a constant. Find the net acceleration of the particle at the instant when it has covered $n$ th fraction of the circle after the beginning of motion.