A 2.1 kg particle-like object moves in a plane with velocity components vx = 30 m/s and vy = 55 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?
kg · m2/s
(b) What is its angular momentum relative to the point (-2.0, -2.0) m at this same moment?
kg · m2/s
HELP!!! I don't know what to do or where to start, except with the angular momentum formula L=r X p=m(r X v. Would the formula l=rmvsintheta also be necessary here??
Looks like we both have angular momentum problems. Im just as far along as you, I only have the formulas... Hopefully someone will know what to do!!!
To calculate the angular momentum, you can use the formula L = r x p, as you noted. The formula L = rmvsinθ represents the magnitude of the angular momentum when the motion is in a plane perpendicular to the position vector r, and θ is the angle between the velocity vector v and the position vector r.
However, in this case, since the motion is in a plane, we can directly use the formula L = r x p, without the need for the magnitude formula.
To calculate the angular momentum relative to the origin, you can follow these steps:
Step 1: Find the position vector r from the origin to the given coordinates (3.0, -4.0) m. The position vector r is given by r = (x, y) = (3.0, -4.0) m.
Step 2: Find the momentum vector p by multiplying the mass m (2.1 kg) with the velocity vector v = (vx, vy) = (30 m/s, 55 m/s). The momentum vector p is given by p = m * v.
Step 3: Take the cross product of the position vector r and the momentum vector p by using the formula L = r x p.
L = r x p
Step 4: Calculate the cross product. Given that the position vector r = (3.0, -4.0) m and the momentum vector p = (2.1 kg) * (30 m/s, 55 m/s), we have:
L = (3.0, -4.0) x (2.1 kg) * (30 m/s, 55 m/s)
Here, x represents the cross product operation.
Step 5: Calculate the cross product by expanding the determinant. The cross product of two vectors can be written as the determinant of the 3x3 matrix:
L = (3.0, -4.0) x (2.1 kg) * (30 m/s, 55 m/s)
= | i j k |
| 3.0 -4.0 0 | x (2.1 kg) * (30 m/s, 55 m/s)
Expanding the determinant, we get:
L = i * (-4.0 * (2.1 kg) * (55 m/s)) - j * (3.0 * (2.1 kg) * (30 m/s)) + k * ((3.0 * (2.1 kg) * (55 m/s)) - (-4.0 * (2.1 kg) * (30 m/s)))
Simplifying the expression, we have:
L = -462 i + 189 j + 221.4 k
Step 6: The result of the cross product is the angular momentum vector L. To find the magnitude of the angular momentum, you can take the magnitude of the vector L using the formula:
|L| = √(Lx^2 + Ly^2 + Lz^2)
In this case, the magnitude is given by:
|L| = √((-462)^2 + (189)^2 + (221.4)^2) kg · m²/s
This will give you the angular momentum relative to the origin.
To calculate the angular momentum relative to the point (-2.0, -2.0) m, follow the same steps as above, but adjust the position vector r to be relative to the new point (-2.0, -2.0) m. Then, repeat the calculation for the cross product and magnitude as shown above.
I hope this helps! Let me know if you need any further clarification.