Question
A particle of mass,m =3.0kg moves with a uniform velocity ,v=(5m/s)i+(3m/s)j. At the time ,t,the particle passes through the point r=(m)i-(3m)j from the origin. Calculate the magnitude and direction of the angular momentum about the origin at time t.
Answers
assume you mean
R = i - 3j
and
V = 5i + 3j
L = m * R cross V
= m *
i j k
1-3 0
5 3 0
= m[ 3 k +15 k] = 18*3 k
or 54 in z (or along k) direction
K = m * [
R = i - 3j
and
V = 5i + 3j
L = m * R cross V
= m *
i j k
1-3 0
5 3 0
= m[ 3 k +15 k] = 18*3 k
or 54 in z (or along k) direction
K = m * [
It really helps but can you please simplify the expression step by step for better understanding....
Related Questions
A particle (m=3*10^-6 g, q=5*10^-6 C) moves in a uniform magnetic field given by (60j) mT. At t=0 th...
A particle (mass = 5.0 g, charge = 40 mC) moves in a region of space where the electric
field is un...
A particle (mass = 5.0 g, charge = 40 mC) moves in a region of space where the electric field is uni...
A particle (mass = 4.0 g, charge = 50 mC) moves in a region of space where the electric field is uni...