Question

the position vector and angular velocity vector of a particle executing uniform circular motion at an instant are 2i^ and 4k^ respectively find its linear velocity at that instant​

Answer 1

Answer:

v = w * r

where * is cross product

hence v = 4 k^ * 2 i^

= 8 j^

Explanation:

Answer 2

Its linear velocity of the particle at that instant is 8 m/s.

Given:

Position vector   = 2i^

Angular velocity = 4k^

To find:

Linear velocity of the particle.

Solution:

Position vector of particle,

r(t) = R cos(ωt)i^ + R sin(ωt)j^

Here,

R= Radius of the circular path.

ω= Angular velocity of the particle.

i^ and j^= Unit vectors of the directions x and y.

Now if we differentiate the position vector,

v(t) = -R ωsin(ωt)i^ + R ωcos(ωt)j^ .

We know,

The magnitude of the velocity is the linear velocity ,

v = |v(t)|

  = √[(-Rωsin(ωt))^2 + (Rωcos(ωt))^2]

Given, Position vector = 2i^,

From this we can say that the distance from the origin = 2 unit

The particle is moving in perpendicular direction of the x-y plane.

Radius of the circular path =2 units.(R)

Angular velocity of the particle  = 4 units/s.( ω)

Now after putting this values ,

v = |v(t)|

  = √[(-2×4sin(ωt))^2 + (2×4cos(ωt))^2]

  = √[64sin^2(ωt) + 64cos^2(ωt)]

  = √(64)

  = 8 m/s

So ,its linear velocity of the particle at that instant is 8 m/s.

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