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) y = bsin(wt). where a,b,w, are constants. determine angular momentum ofthe particle with respect to the origin of the coordinate system.
To determine the angular momentum of the particle with respect to the origin of the coordinate system, we need to find the orbital angular momentum. The formula for orbital angular momentum is given by L = r x p, where r is the position vector and p is the linear momentum.
First, let's find the position vector r of the particle. The position vector r can be written as r = xi + yj, where i and j are unit vectors along the x and y directions, respectively.
Given:
x = acos(wt)
y = bsin(wt)
Substituting these values, we have:
r = (acos(wt))i + (bsin(wt))j
Now, let's find the linear momentum p of the particle. The linear momentum p can be written as p = mv, where m is the mass of the particle and v is the velocity.
The velocity of the particle can be found by taking the derivative of the position vector with respect to time:
v = dr/dt
Differentiating both x and y with respect to time, we get:
dx/dt = -awasin(wt)
dy/dt = bcos(wt)
Substituting these values, we have:
v = (-awasin(wt))i + (bcos(wt))j
Now, we can calculate the linear momentum p:
p = mv = m[(-awasin(wt))i + (bcos(wt))j]
Next, let's calculate the cross product of r and p:
r x p = |i j k |
|acos(wt) bsin(wt) 0 |
| -awasin(wt) bcos(wt) 0 |
Expanding the determinant, we get:
r x p = 0i - 0j + [(acos(wt))(bcos(wt))]k - [(bsin(wt))(-awasin(wt))]k
= [(acos(wt))(bcos(wt))]k + [(bsin(wt))(-awasin(wt))]k
Now, let's simplify the expression:
r x p = [(acos(wt))(bcos(wt)) + (bsin(wt))(-awasin(wt))]k
Finally, the angular momentum L of the particle with respect to the origin is given by:
L = r x p = [(acos(wt))(bcos(wt)) + (bsin(wt))(-awasin(wt))]k
Therefore, the formula for angular momentum of the particle with respect to the origin of the coordinate system is L = [(acos(wt))(bcos(wt)) + (bsin(wt))(-awasin(wt))]k.
To determine the angular momentum of the particle with respect to the origin of the coordinate system, we need to determine the momentum and position vectors of the particle and then calculate their cross product.
The momentum vector of the particle is given by:
p = mv
where m is the mass of the particle and v is its velocity vector.
The velocity vector can be obtained by taking the time derivative of the position vector:
v = d/dt(r) = d/dt(x)i + d/dt(y)j
= -awsin(wt)i + bcos(wt)j
where i and j are the unit vectors along the x and y axes, respectively.
Now we have the momentum vector:
p = m(-awsin(wt)i + bcos(wt)j)
= -mawsin(wt)i + mbcos(wt)j
To calculate the cross product of the position and momentum vectors, we arrange them in determinant form:
L = r x p = |i j k|
|ax ay az|
|-mawsin(wt) mbcos(wt) 0|
Expanding this determinant, we get:
L = (ay*mawsin(wt) - az*mbcos(wt))i - (ax*mawsin(wt) - az*mbcos(wt))j + (ax*mbcos(wt) - ay*mbcos(wt))k
Since the position vector is in the x-y plane, the z-component of the angular momentum is zero:
Lz = ax*mbcos(wt) - ay*mbcos(wt) = (ax - ay)mbcos(wt)
Therefore, the angular momentum of the particle with respect to the origin of the coordinate system is given by:
L = (ay*mawsin(wt) - az*mbcos(wt))i - (ax*mawsin(wt) - az*mbcos(wt))j + (ax - ay)mbcos(wt)k