a particle of mass m moves in a x-y plane. The coordinates of the particle at any instant are given by x = acos(wt) and y = bsin(wt). where a,b,w, are constant. Determine the angular momentum of the particle with respect to the origin of the coordinate system.
To determine the angular momentum of the particle with respect to the origin, we need to calculate the cross product of the position vector and the linear momentum vector of the particle.
The position vector of the particle at any time t is given by r = xi + yj, where i and j are the unit vectors in the x and y directions, respectively.
So, the position vector r = x(t)i + y(t)j = (acos(wt))i + (bsin(wt))j.
The linear momentum vector p is defined as the product of the mass m and the velocity vector v. The particle's velocity vector v can be obtained by differentiating the position vector with respect to time.
dv/dt = dx/dt i + dy/dt j
= -awsin(wt) i + bwcos(wt) j
The velocity vector is v = -awsin(wt) i + bwcos(wt) j.
Now, we can calculate the cross product of the position vector r and the linear momentum vector p:
L = r × p
L = (acos(wt)i + bsin(wt)j) × (-awsin(wt)i + bwcos(wt)j)
Expanding the cross product, we have:
L = (acos(wt) * -awsin(wt) - bsin(wt) * bwcos(wt)) k
L = -a^2w*sin^2(wt) - b^2w*cos^2(wt) k
Therefore, the angular momentum of the particle with respect to the origin is given by:
L = -a^2w*sin^2(wt) - b^2w*cos^2(wt) k