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Systems of Particles and Rotational Motion

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Question : 33
Total: 33
Separation of motion of a system of particles into motion of the centre of mass and motion about the centre of mass
(a) Show
→
p
=
→
p
i′+mi
→
v

where pi is the momentum of the ith particle (of mass mi) and pi′=mivi′. Note vl′ is the velocity of the i ‌th particle relative to the centre of mass.
Also, prove using the definition of the centre of mass ∑
→
p
i′=0
(b) Show K=K′+
1
2
MV2
where K is the total kinetic energy of the system of particles, K′ is thetotal kinetic energy of the system when the particle velocities are takenwith respect to the centre of mass and MV2∕2 is the kinetic energy of thetranslation of the system as a whole (i.e. of the centre of mass motion of thesystem).
(c) Show
→
L
=
→
L
+
→
R
×M
→
v

where
→
L
′=∑
→
r
i′×
→
p
i′ is the angular momentum of the system about the centre ofmass with velocities taken relative to the centre of mass. Remember
→
r
i′=
→
r
i−
→
R
; rest of the notation is the standard notation used in the chapter. Note
→
L
′ and M
→
R
×
→
V
 can be said to be angular momenta, respectively, about and of thecentre of mass of the system of particles.
(d) Show
d
→
L
′
dt
=∑
→
r
i′×
d
→
p
′
dt

Further, show that
d
→
L
′
dt
=τ′ext where τ′ext is the sum of all external torques acting on the system about the centre of mass.
(Hint : Use the definition of centre of mass and Newton’s Third law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:  
(a) Consider a system of i moving particles.
Mass of ith particle =mi and velocity of the ith particle =
→
v
i
Hence momentum of i ‌th particle
→
p
i=mi
→
v
i .
Velocity of the centre of mass =
→
V

Velocity of the i ‌th particle with respect to the centre of mass of the system is
→
v
i′=
→
v
i−V or mi
→
v
i′=mi
→
v
i−mi
→
V

∴
→
p
i′=
→
p
i−mi
→
V
 (∵
→
p
i′=mi
→
v
i′)
or
→
p
i=
→
p
i′+mi
→
V

Hence proved.
Now, ∑
→
p
i′=∑mi
→
v
i′=∑mi
d
→
r
i′
dt

As per the definition of centre of mass, ∑mi
→
r
i′=0
∴∑mi
d
→
r
i′
dt
=0‌or‌∑
→
p
i′=0
(b) K.E. of a system consists of two parts translational K.E. ( Kt ) and rotational K.E. (K′) i.e. K.E. of motion of C.M. (
1
2
mv2) and K.E. of rotational motion about the C.M. of the system of particle (K′), thus total K.E. of the systemis given by
K=
1
2
mv2+
1
2
Iω2
=
1
2
mv2+K′=K′+
1
2
mv2
(c) As
→
p
i=
→
p
i′+mi
→
V

∴∑
→
r
i×
→
p
i=∑
→
r
i×
→
p
i′+∑
→
r
i×mi
→
V
or
→
L
=∑(
→
r
i′+
→
R
)×
→
p
i′+∑(
→
r
i′+
→
R
)×mi
→
V
=∑
→
r
′i×
→
p
′i+∑
→
R
×
→
p
i′+∑
→
r
′i×mi
→
V
+∑
→
R
×mi
→
V
=L′+∑
→
R
×
→
p
i′+∑
→
r
′i×mi
→
V
+∑
→
R
×mi
→
V
Now,
→
R
×∑
→
p
i′=0,(
∑
i
→
r
i′)×M
→
V
=0 and ‌∑mi=M∴
→
L
=
→
L
′+
→
R
×M
→
V

(d) As
→
L
i′=
→
r
i′×
→
p
i′
∴ rate of change of the angular momentum of a particle is given by
dLi′
dt
=
d
dt
(ri′×pi′)
=ri′×
dpi′
dt
+
dri′
dt
×pi′
=ri′×
dpi′
dt
+vi′×pi′
=ri′×
dpi′
dt
(∵vi′×pi′=0)
If L′ be the total angular momentum of the system, then
L=∑Li′
∴
dLi′
dt
=∑
dLi′
dt
=∑(ri′×
dpi′
dt
)...(i)
Hence proved.Also we know that if τ′ext be the total external torque acting on the system, then
τ′ext=∑(ri′×
dpi′
dt
)
=ri′×∑
dpi′
dt
=ri′×F′ext...(ii)
(∵ external forces always appear in pairs and then cancel each other).
∴ From (i) and (ii), we get
dLi′
dt
= τ′ext
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