Question

Momentum
enersy
energy wil
particles
time;
E
dr x
dp
(rx p)
dt dt
хр+rx
dt
dr
=vx mv + rxF
dt
But vx mv = m(v x v) = 0
dJ
=rxF
dp - F
F
dt
B
about
...(7.59)
rota
By the definition of torque, t=rxF
dJ
dt
Which means the rate of change of angular
momentum is equal to the torque applied on it.
This statement is equivalent to Newton's second law in
dp
linear motion F
dt
ene
m
k
7.22. Relation between Torque and Angular
Differentiating equation (7.58) with respect to​

Answer 1

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Explanation:

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