Question

A particle is moving in a circle with constant angular velocity a . If r vector is position vector of particle then show by vector method, that the acceleration is –w^2 r .

please solve it​

Answer 1

Answer:

Solution

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Correct option is D)

Position :

R

=4sin(2πt)

i

^

+4cos(2πt)

j

^

Velocity:

v

=4×2πcos(2πt)

i

^

−4×2πsin(2πt)

j

^

Acceleration: a=−4(2π)

2

sin(2πt)

i

^

−4(2π)

2

cos(2πt)

j

^

=−(2π)

2

R

∴ acceleration is along −

R

Magnitude of velocity: ∣

v

∣=

(4×2πcos(2πt))

2

+(−4×2πsin(2πt))

2

=

(8π)

2

sin

2

(2πt)+cos

2

(2πt)

=8π

Magnitude of acceleration : ∣

a

∣=∣−(2π)

2

R

∣=4π

2

∣

R

∣=16π

2

=

4

(8π)

2

=

R

v

2

R

x

=x=4sin(2πt)

R

y

=y=4cos(2πt)

R

x

2

+R

y

2

=x

2

+y

2

=4

2

⇒ the path of the particle is a circle of radius 4.

Hence, option(D) is wrong.

Explanation:

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