At a certain instant, the position of a 0.25kg body is r= (2 i-2j)m and its velocity is v=(-5i+5k)m/s with a net force f=4jN is acting on it. Then what is the angular momentum of a body about the origin?
The angular momentum of a body about the origin is given by L=r x p, where r is the position vector and p is the momentum vector.
First, we need to find the momentum vector. The momentum of the body is given by p=mv, where m is the mass and v is the velocity vector.
Here, m=0.25kg and v=(-5i+5k)m/s. Therefore, p=0.25*(-5i+5k)=-1.25i+1.25k Ns.
Next, we need to find the position vector. The position vector of the body is r=2i-2j m.
Now, we can calculate the cross product of r and p:
r x p = |i j k|
|2 -2 0|
|-1.25 0 1.25|
= (2*1.25)i + (2*1.25)j + (-2*(-1.25))k
= 2.5i + 2.5j + 2.5k
Therefore, the angular momentum of the body about the origin is 2.5i+2.5j+2.5k Ns.