A particle moves in a plane according to the law v=v0*i+b*omega*cosomega*t*jIf the particle is at origin at time t=0, determine the equation of the particle path y=f(x) and the distance from the...

A particle moves in a plane according to the law v=v0*i+b*omega*cosomega*t*j

If the particle is at origin at time t=0, determine the equation of the particle path y=f(x) and the distance from the origin at time t=3pi/2*omega

2 Answers | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The formula for the vector of velocity is:

v = vx*i + vy*j 

We'll identify the projections of velocity form the formula v=v0*i+b*omega*cosomega*t*j:

- x axis projection:

vx =  dx/dt = v0

dx = v0dt

Int dx = Int v0dt

x = v0*t + C

t = x/v0 (1)

- y axis projection:

vy = dy/dt = b*omega*cos(omega)*t

dy = b*omega*cos(omega)*t dt

Int dy = Int b*omega*cos(omega)*t dt

y = b*t*sin(omega) + C (2)

We'll substitute (1) in (2):

y = b*(x/v0)*sin(omega)

If t = 3pi/2*omega => x = 3pi*v0/2*omega

y = b*(3pi*v0/2*omega*v0)*sin(omega)

y = b*sin(3pi/2)

y = b*(-1)

y = -b

The distance of the particle from the origin is:

r = sqrt(x^2 + y^2)

r = sqrt[9(pi*v0)^2/4*omega^2 + b^2]

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The velocity components at time t is vo  in the direction of i or X and

bwcoswt in the direction of j or Y.

So dx/dt = v0

dy/ddt = (bwcosw)t

Therefore x = Int v0 dt = t=v0t+ C1. But since x is at origin at time t0, 0+vt+ C1 . So C1 = 0.

x = v0t.........(1).

dy/dt = (bwcosw)t.

 y = Int(bwcosw)tdt = (bwcosw)t^2/2 +C2. Applying the initial condtion that the particle is at origin, we find C2 = 0.

 So y = (bwcosw)*t^2/2..........(2). 

From(1):  t = x/v0.   Substitute t = x/v0 in (2):

y = (bwcosw)(x/v0)^2 is the relation between x and y.

 At time t = 3pi/2, the distance of the particle from the origin is sart(x^2+y^2) = sqrt{(v*t)^2 +(bwcosw)t^4} = (3pi v0 /2)*{1+(9/4pi^2)*bwcos^2w}

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question