MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relative Velocity in 2D

Find the magn...

Question

Find the magnitude of displacement of a particle moving in a circle of radius $r$ of constant angular sped $w$ as a function of time $t$

Open in App

Solution

Particle is at the extreme left of diameter
$x = a c o s ϕ y = a s i n ϕ r = \sqrt{{(a + a c o s ϕ)}^{2} + {(a s i n ϕ)}^{2}} = a \sqrt{1 + {cos}^{2} ϕ + 2 c o s ϕ + {sin}^{2} ϕ} = a \sqrt{2} \sqrt{2} c o s \frac{ϕ}{2} = 2 a c o s (\frac{ϕ}{2}) m a g n i t u d e o f d i s p l a c e m e n t w h i c h i s r = 2 a c o s (\frac{ϕ}{2}) = 2 a c o s (\frac{ω t}{2})$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The magnitude of displacement of a particle moving in a circle of radius

a

with constant angular speed

ω

varies with time

t

Q. A particle moves along a circle of radius R with a constant angular speed

ω

its displacement (only magnitude) in time t will be

Q. The magnitude of the displacement of a particle moving in a circle of radius

a

with constant angular speed

ω

varies with time

t

Q. A particle is moving in a circle of radius

R

with constant speed. The time period of the particle is

T

t = \frac{T}{6}

, the magnitude of average velocity of the particle is

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Relative Motion in 2D

PHYSICS

Watch in App

explore_more_icon

Explore more

Relative Velocity in 2D

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon