Posted by COFFEE on Friday, March 30, 2007 at 4:25am.
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!
For Further Reading
* Physics - bobpursley, Friday, March 30, 2007 at 7:52am
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.
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thank you, i understand that part.
..but if the origin was changed, in this case in part b: (-2.0, -2.0) m
how would the 'Angular momentum= R x mV' equation change? is the linear momentum the one affected?
R changes: New R is R-Ro where Ro is the new point for the origin.
You now have M=(R-Ro)xV
thanks! got it :)
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To find the angular momentum relative to the origin, you need to multiply the position vector (r) by the linear momentum (mv). The position vector is the vector from the origin to the point where the object is located. In this case, the position vector is (3.0i - 4.0j) m.
The linear momentum can be calculated by multiplying the mass of the object (1.1 kg) by the velocity vector (v). The velocity vector has the x-component Vx = 30 m/s and the y-component Vy = 90 m/s. Therefore, the linear momentum is (1.1 kg)(30i + 90j) kg*m/s.
Now you can calculate the angular momentum relative to the origin:
Angular momentum = r x p = (3.0i - 4.0j) m x (1.1 kg)(30i + 90j) kg*m/s.
To do the cross product, you need to use the determinant method:
(3.0i - 4.0j) m x (1.1 kg)(30i + 90j) kg*m/s
= (3.0)(1.1)(i x i) + (3.0)(1.1)(i x j) + (-4.0)(1.1)(j x i) + (-4.0)(1.1)(j x j) m^2*kg/s.
Since i x i = j x j = 0 and i x j = -j x i = k, you can simplify the equation:
= (3.0)(1.1)(i x j - j x i) k m^2*kg/s
= (3.0)(1.1)(-2k) m^2*kg/s
= -6.6k m^2*kg/s.
Therefore, the angular momentum relative to the origin is -6.6 kg*m^2/s in the k-direction.
To find the angular momentum relative to the point (-2.0, -2.0) m, you need to calculate the new position vector (R) relative to the new origin. The new position vector is (3.0 - (-2.0))i - (4.0 - (-2.0))j m = 5.0i - 6.0j m.
Using the same linear momentum (mv), the new angular momentum relative to the new origin is:
Angular momentum = R x p = (5.0i - 6.0j) m x (1.1 kg)(30i + 90j) kg*m/s.
Again, you can use the determinant method to do the cross product:
(5.0i - 6.0j) m x (1.1 kg)(30i + 90j) kg*m/s
= (5.0)(1.1)(i x i) + (5.0)(1.1)(i x j) + (-6.0)(1.1)(j x i) + (-6.0)(1.1)(j x j) m^2*kg/s.
Simplifying the equation:
= (5.0)(1.1)(i x j - j x i) k m^2*kg/s
= (5.0)(1.1)(-2k) m^2*kg/s
= -11.0k m^2*kg/s.
Therefore, the angular momentum relative to the point (-2.0, -2.0) m is -11.0 kg*m^2/s in the k-direction.