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Systems of Particles and Rotational Motion

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Question : 33 of 33
Marks: +1, -0
Separation of motion of a system of particles into motion of the centre of mass and motion about the centre of mass
(a) Show p⃗=p⃗i′+miv⃗\vec{p} = \vec{p}_i' + m_i \vec{v}
where pip_{i} is the momentum of the ithi^{\text{th}} particle (of mass mi)m_{i} ) and pi′=mivi′.p_{i}'=m_{i} v_{i}' . Note vl′v_{l}' is the velocity of the ithi^{\text{th}} particle relative to the centre of mass.
Also, prove using the definition of the centre of mass ∑p⃗i′=0\sum \vec{p}_i' = 0
(b) Show K=K′+12MV2K = K' + \frac{1}{2} M V^{2}
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 MV22\frac{M V^{2}}{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 L⃗=L⃗+R⃗×Mv⃗\vec{L} = \vec{L} + \vec{R} \times M \vec{v}
where L⃗′=∑r⃗i′×p⃗i′\vec{L}' = \sum \vec{r}_i' \times \vec{p}_i' is the angular momentum of the system about the centre ofmass with velocities taken relative to the centre of mass. Remember r⃗i′=r⃗i−R⃗;\vec{r}_i' = \vec{r}_i - \vec{R}; rest of the notation is the standard notation used in the chapter. Note L⃗′\vec{L}' and MR⃗×V⃗M \vec{R} \times \vec{V} can be said to be angular momenta, respectively, about and of thecentre of mass of the system of particles.
(d) Show dL⃗′dt=∑r⃗i′×dp⃗′dt\frac{d \vec{L}'}{d t} = \sum \vec{r}_i' \times \frac{d \vec{p}'}{d t}
Further, show that dL⃗′dt=τext′\frac{d \vec{L}'}{d t} = \tau'_{\mathrm{ext}} where τext′\tau'_{\mathrm{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 ithi^{\text{th}} particle =mi= m_{i} and velocity of the ithi^{\text{th}} particle =v⃗i= \vec{v}_{i}
Hence momentum of ithi^{\text{th}} particle p⃗i=miv⃗i\vec{p}_i = m_{i} \vec{v}_i .
Velocity of the centre of mass =V⃗= \vec{V}
Velocity of the ithi^{\text{th}} particle with respect to the centre of mass of the system is
v⃗i′=v⃗i−V\vec{v}_i' = \vec{v}_i - V or miv⃗i′=miv⃗i−miV⃗m_{i} \vec{v}_i' = m_{i} \vec{v}_i - m_{i} \vec{V}
∴p⃗i′=p⃗i−miV⃗ (∵p⃗i′=miv⃗i′)\therefore \vec{p}_i' = \vec{p}_i - m_{i} \vec{V} \ (\because \vec{p}_i' = m_{i} \vec{v}_i')
or p⃗i=p⃗i′+miV⃗\vec{p}_i = \vec{p}_i' + m_{i} \vec{V}
Hence proved.
Now, ∑p⃗i′=∑miv⃗i′=∑midr⃗i′dt\sum \vec{p}_i' = \sum m_i \vec{v}_i' = \sum m_i \frac{d \vec{r}_i'}{d t}
As per the definition of centre of mass, ∑mir⃗i′=0\sum m_i \vec{r}_i' = 0
∴∑midr⃗i′dt=0or∑p⃗i′=0\therefore \sum m_i \frac{d \vec{r}_i'}{d t} = 0 \quad \text{or} \quad \sum \vec{p}_i' = 0
(b) K.E. of a system consists of two parts translational K.E. ( KtK_t ) and rotational K.E. (K′) i.e. K.E. of motion of C.M. (12mv2)\left(\frac{1}{2} m v^{2}\right) 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=12mv2+12Iω2K = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}
=12mv2+K′=K′+12mv2= \frac{1}{2} m v^{2} + K' = K' + \frac{1}{2} m v^{2}
(c) As p⃗i=p⃗i′+miV⃗\vec{p}_i = \vec{p}_i' + m_{i} \vec{V}
∴∑r⃗i×p⃗i=∑r⃗i×p⃗i′+∑r⃗i×miV⃗\therefore \sum \vec{r}_i \times \vec{p}_i = \sum \vec{r}_i \times \vec{p}_i' + \sum \vec{r}_i \times m_{i} \vec{V}
or L⃗=∑(r⃗i′+R⃗)×p⃗i′+∑(r⃗i′+R⃗)×miV⃗\vec{L} = \sum (\vec{r}_i' + \vec{R}) \times \vec{p}_i' + \sum (\vec{r}_i' + \vec{R}) \times m_i \vec{V}
=∑r⃗i′×p⃗i′+∑R⃗×p⃗i′+∑r⃗i′×miV⃗+∑R⃗×miV⃗= \sum \vec{r}_i' \times \vec{p}_i' + \sum \vec{R} \times \vec{p}_i' + \sum \vec{r}_i' \times m_i \vec{V} + \sum \vec{R} \times m_i \vec{V}
=L⃗′+∑R⃗×p⃗i′+∑r⃗i′×miV⃗+∑R⃗×miV⃗= \vec{L}' + \sum \vec{R} \times \vec{p}_i' + \sum \vec{r}_i' \times m_i \vec{V} + \sum \vec{R} \times m_i \vec{V}
Now, R⃗×∑p⃗i′=0,(∑ir⃗i′)×MV⃗=0\vec{R} \times \sum \vec{p}_i' = 0, \quad \left( \sum_{i} \vec{r}_i' \right) \times M \vec{V} = 0 and ∑mi=M∴L⃗=L⃗′+R⃗×MV⃗\sum m_i = M \quad \therefore \vec{L} = \vec{L}' + \vec{R} \times M \vec{V}
(d) As L⃗i′=r⃗i′×p⃗i′\vec{L}_i' = \vec{r}_i' \times \vec{p}_i'
∴ rate of change of the angular momentum of a particle is given by
dLi′dt=ddt(ri′×pi′)\frac{d L_i'}{d t} = \frac{d}{d t} (r_i' \times p_i')
=ri′×dpi′dt+dri′dt×pi′= r_i' \times \frac{d p_i'}{d t} + \frac{d r_i'}{d t} \times p_i'
=ri′×dpi′dt+vi′×pi′= r_i' \times \frac{d p_i'}{d t} + v_i' \times p_i'
=ri′×dpi′dt(∵vi′×pi′=0)= r_i' \times \frac{d p_i'}{d t} \quad (\because v_i' \times p_i' = 0)
If L′L' be the total angular momentum of the system, then
L=∑Li′L = \sum L_i'
∴dLi′dt=∑dLi′dt=∑(ri′×dpi′dt)...(i)\therefore \frac{d L_i'}{d t} = \sum \frac{d L_i'}{d t} = \sum \left(r_i' \times \frac{d p_i'}{d t}\right) \quad \text{...(i)}
Hence proved.Also we know that if τext′\tau'_{\mathrm{ext}} be the total external torque acting on the system, then
τext′=∑(ri′×dpi′dt)\tau'_{\mathrm{ext}} = \sum \left(r_i' \times \frac{d p_i'}{d t}\right)
=ri′×∑dpi′dt=ri′×Fext′...(ii)= r_i' \times \sum \frac{d p_i'}{d t} = r_i' \times F'_{\mathrm{ext}} \quad \text{...(ii)}
(∵ external forces always appear in pairs and then cancel each other).
∴ From (i) and (ii), we get
dLi′dt=τext′\frac{d L_i'}{d t} = \tau'_{\mathrm{ext}}
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