A 1.1 kg particle-like object moves in a plane with velocity components Vx = 30 m/s and Vy = 90 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*m^2/s k
(b) What is its angular momentum relative to the point (-2.0, -2.0) m at this same moment?
_____ kg*m^2/s k
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i drew a points on a flat xy plane.. but im not sure if i drew it correctly. i have the Vx component towards the positive X direction and the Vy towards the negative Y direction.
angular momentum = r x p
r x p = m (r x v)
......= 1.1 (r x 30)
......= 1.1r x 30
-33 = 1.1r
-30 = r
.. im stuck.. help!
1. Angular momentum= R x mV Where mV is the linear momentum, x is the cross product, and R is the position vector.
Angular momenum= m(3i-4j)x(30i+90j)
= m270k + m 120k
= m*390 k
check my thinking.
Why did you do the Vy in the negative y direction?
To calculate the angular momentum of the particle-like object, we need to use the formula L = r x p, where L is the angular momentum, r is the position vector, and p is the linear momentum.
(a) Angular momentum relative to the origin at (3.0, -4.0) m:
Given:
m = 1.1 kg (mass of the object)
Vx = 30 m/s (velocity component in the x-direction)
Vy = 90 m/s (velocity component in the y-direction)
r = (3.0i - 4.0j) m (position vector)
To find the linear momentum p, we can use the formula p = m * V:
p = (m * Vx)i + (m * Vy)j
= (1.1 kg * 30 m/s)i + (1.1 kg * 90 m/s)j
= 33i + 99j kg*m/s
Now, we can calculate the cross product of r and p:
r x p = (3.0i - 4.0j) x (33i + 99j)
= (3 * 99 - (-4 * 33))k
= 297k + 132k
= 429k kg*m^2/s
So, the angular momentum relative to the origin is 429 kg*m^2/s in the k-direction.
(b) Angular momentum relative to (-2.0, -2.0) m:
To find the position vector relative to (-2.0, -2.0) m, we subtract the coordinates of the point from the position vector:
r' = r - (-2.0i - 2.0j) m
= (3.0i - 4.0j) - (-2.0i - 2.0j)
= 5.0i - 2.0j m
Using the same linear momentum p as calculated in part (a):
Now, we can calculate the cross product of r' and p:
r' x p = (5.0i - 2.0j) x (33i + 99j)
= (5 * 99 - (-2 * 33))k
= 495k + 66k
= 561k kg*m^2/s
So, the angular momentum relative to the point (-2.0, -2.0) m is 561 kg*m^2/s in the k-direction.
Remember to carefully check the signs and units in your calculations, as they can affect the final answer.