A 1.8 kg particle-like object moves in a plane with velocity components vx = 20 m/s and vy = 50 m/s as it passes through the point with (x, y) coordinates of (3.0, -4.0) m.
(a) What is its angular momentum relative to the origin at this moment?
1 kg · m2/s
(b) What is its angular momentum relative to the point (-2.0, -2.0) m at this same moment?
2 kg · m2/s
1(a)
r = (3 -4); p = 1.8x(20 30)
r X p = 1.8[ 3x50 - (-4x20)] = 1.8x(150 + 80)
= 1.8x230 = 414 kg m2/sec
1(b)
r = (3 - (-2), -4 - (-2)) = ( 5 -2)
similarly
r X p = 522 kg m2/sec
(a) Well, it sounds like this particle-like object is doing quite the dance in the plane, doesn't it? To calculate its angular momentum relative to the origin, we need to calculate the cross product of its position vector and its momentum vector. The position vector is given by (x, y) = (3.0, -4.0) m, and the momentum vector is given by (px, py) = (1.8 kg * 20 m/s, 1.8 kg * 50 m/s).
So, let's calculate this amazing maneuver. The cross product is given by the formula L = r x p, where L is the angular momentum, r is the position vector, and p is the momentum vector. Plugging in the numbers, we get:
L = (3.0, -4.0) m x (1.8 kg * 20 m/s, 1.8 kg * 50 m/s)
L = (3.0 * (1.8 kg * 50 m/s)) - (4.0 * (1.8 kg * 20 m/s))
L = 90 kg*m²/s - 144 kg*m²/s
L = -54 kg*m²/s
Voila! The angular momentum relative to the origin is -54 kg*m²/s. But don't worry, this particle-like object isn't feeling down, it's just spinning things up!
(b) Now, let's twist things a bit and calculate the angular momentum relative to the point (-2.0, -2.0) m. To do this, we need to calculate the position vector relative to that point.
The position vector relative to (-2.0, -2.0) m is given by (x, y) - (-2.0, -2.0) m = (3.0, -4.0) m - (-2.0, -2.0) m
Performing the subtraction, we get (3.0 + 2.0, -4.0 + 2.0) m = (5.0, -2.0) m
Now that we have the relative position vector, we can calculate the angular momentum using the same formula as before: L = r x p, where r is the new relative position vector and p is the momentum vector.
Plugging in the numbers, we get:
L = (5.0, -2.0) m x (1.8 kg * 20 m/s, 1.8 kg * 50 m/s)
L = (5.0 * (1.8 kg * 50 m/s)) - (-2.0 * (1.8 kg * 20 m/s))
L = 90 kg*m²/s - (-72 kg*m²/s)
L = 162 kg*m²/s
Voila! The angular momentum relative to the point (-2.0, -2.0) m is 162 kg*m²/s. Quite the spinning shenanigans, wouldn't you say?
To calculate the angular momentum of the particle-like object, we can use the formula:
L = r x p
where L represents the angular momentum, r is the position vector relative to the point of reference, and p is the linear momentum.
(a) To find the angular momentum relative to the origin, we can use the coordinates (x, y) = (3.0, -4.0) m and the velocity components (vx, vy) = (20 m/s, 50 m/s).
First, we calculate the linear momentum:
p = m * v
where m is the mass of the object and v is its velocity. In this case, the mass (m) is 1.8 kg.
p = 1.8 kg * sqrt((20 m/s)^2 + (50 m/s)^2)
p = 1.8 kg * sqrt(400 m^2/s^2 + 2500 m^2/s^2)
p = 1.8 kg * sqrt(2900 m^2/s^2)
p ≈ 1.8 kg * 53.8516 m/s
p ≈ 96.9329 kg * m/s
Next, we calculate the position vector:
r = (x, y)
r = (3.0 m, -4.0 m)
Finally, we calculate the angular momentum relative to the origin:
L = r x p
L = (3.0 m, -4.0 m) x 96.9329 kg * m/s
L = 96.9329 kg * m/s * (-4.0 m)
L = -387.7316 kg * m^2/s
Therefore, the angular momentum relative to the origin is approximately -387.7316 kg · m^2/s.
(b) To find the angular momentum relative to the point (-2.0, -2.0) m, we need to calculate the new position vector relative to this point.
The new position vector is:
r_new = r - r_ref
where r_ref represents the reference point (-2.0, -2.0) m.
r_new = (x_new, y_new) = (3.0 m, -4.0 m) - (-2.0 m, -2.0 m)
r_new = (5.0 m, -2.0 m)
Now, we can calculate the angular momentum relative to the new reference point:
L = r_new x p
L = (5.0 m, -2.0 m) x 96.9329 kg * m/s
L = 96.9329 kg * m/s * (-2.0 m)
L = -193.8658 kg * m^2/s
Therefore, the angular momentum relative to the point (-2.0, -2.0) m is approximately -193.8658 kg · m^2/s.
To find the angular momentum of the particle-like object, we can use the formula for angular momentum:
L = r × p
where L is the angular momentum, r is the displacement vector from the reference point to the object, and p is the momentum vector of the object.
(a) To find the angular momentum relative to the origin (0, 0), we can calculate the displacement vector r and the momentum vector p.
The displacement vector r is given by subtracting the coordinates of the reference point (0, 0) from the coordinates of the object (3.0, -4.0):
r = (3.0, -4.0) - (0, 0) = (3.0, -4.0)
The momentum vector p is given by multiplying the mass of the object (1.8 kg) by its velocity vector (vx, vy):
p = m * v = (1.8 kg) * (20 m/s, 50 m/s) = (36 kg·m/s, 90 kg·m/s)
Now, we can calculate the angular momentum by taking the cross product of the displacement vector and the momentum vector:
L = r × p = (3.0, -4.0) × (36, 90)
Using the cross product formula, we can calculate the magnitude of the cross product as:
L = |r × p| = |(3.0 * 90) - (-4.0 * 36)| = |(270) + (144)| = 414 kg·m²/s
Therefore, the angular momentum of the object relative to the origin at this moment is 414 kg·m²/s.
(b) To find the angular momentum relative to the point (-2.0, -2.0), we use the same formula as above.
The displacement vector r is given by subtracting the coordinates of the reference point (-2.0, -2.0) from the coordinates of the object (3.0, -4.0):
r = (3.0, -4.0) - (-2.0, -2.0) = (5.0, -2.0)
The momentum vector p remains the same as in part (a):
p = (36 kg·m/s, 90 kg·m/s)
Now, we can calculate the angular momentum:
L = r × p = (5.0, -2.0) × (36, 90)
Using the cross product formula, we can calculate the magnitude of the cross product as:
L = |r × p| = |(5.0 * 90) - (-2.0 * 36)| = |(450) + (72| = 522 kg·m²/s
Therefore, the angular momentum of the object relative to the point (-2.0, -2.0) at this moment is 522 kg·m²/s.