Answer to Question #198788 in Quantum Mechanics for arsene mihigo

Question

14) A particle of mass m moves according to

2 3

x = x + at , y = bt , z = ct , where a, b and c are

constants.(a)Find the angular momentum L

r

at any time. (b) Find the force F

r

, and from it the

torque τ

r

acting on the particle. Verify that these quantities satisfy

dL

r F

dt

= × =τ

Answer 1

Let us consider the diagram below

Solution

1. Particle one :

$r_1=2.0mi+1.0mj,p_1=2.0kg(4.0m/sj)=8.0kg.m/sj$

$l_r=r_1 \times p_1=-16.0kg.m^2/sk.$

particle 2:

$r_2=4.0mi+1.0mj,p_2=4.0kg(5.0m/sj)=20.0kg.m/si$

$l_2=r_2 \times p_2=-20.0kg.m^2/sk.$

particle 3:

$r_3=2.0mi+2.0mj,p_3=1.0kg(3.0m/sj)=3.0kg.m/si$

$l_3=r_3\times p_3=-6.0kg.m^2/sk.$

we add the individual angular moments to find the total about the origin:

$l_r=l_1+l_2+l_3=-30kg.m^2/sk$

2.The individual forces and lever arms are

$r_{1\perp}=1.0mj, F_1=-6.0N_i,\tau_1=6.0N.mk$

$r_{2\perp}=4.0mi, F_2=10.0N_j,\tau_2=40.0N.mk$

$r_{3\perp}=2.0mi, F_3=-8.0N_j,\tau_3=-16.0N.mk$

Therefore:

$\sum_i\tau_i=\tau_1+\tau_2+\tau_3=30N.mk.$