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Chapter 5: Problem 8

The position vector of the centre of mass of a system of particles in any inertial frame is defined by \(\mathbf{r}_{\mathrm{CM}}=\Sigma m r / \Sigma m\). If the particles suffer only collision forces, prove that \(\dot{\boldsymbol{r}}_{\mathrm{CM}}=u_{\mathrm{CM}}(\cdot \equiv \mathrm{d} / \mathrm{d} t)\); i.e. the centre of mass moves with the velocity of the CM frame. [Hint: \(\Sigma m\), \Sigmami are constant; \Sigmamir is zero between collisions, and at any collision we can factor out the r of the participating particles: \(r \Sigma \dot{m}=0 .]\)

Short Answer

Expert verified
Question: Prove that the velocity of the center of mass of a system of particles suffering only collision forces is equal to the velocity of the center of mass frame. Answer: Differentiating the given position vector of the center of mass with respect to time and using the hint provided, we find that the velocity of the center of mass (\(\dot{\boldsymbol{r}}_{\mathrm{CM}}\)) is equal to the velocity of the center of mass frame (\(u_{\mathrm{CM}}\)). This demonstrates the equality between the two velocities.

Step by step solution

01

Understand the position vector of the center of mass formula

We are given the position vector for the center of mass (\(\mathbf{r}_{\mathrm{CM}}\)) as \(\mathbf{r}_{\mathrm{CM}}=\Sigma m r / \Sigma m\). This equation means that the position vector of the center of mass is the sum of the product of mass (\(m\)) and position vector (\(r\)) of each particle divided by the total mass of the system.
02

Differentiating the position vector of the center of mass

To find \(\dot{\boldsymbol{r}}_{\mathrm{CM}}\), we need to differentiate the position vector of the center of mass with respect to time. The formula is \(\mathbf{r}_{\mathrm{CM}}=\Sigma m r / \Sigma m\), so differentiating both sides with respect to time, we get: \[ \dot{\mathbf{r}}_{\mathrm{CM}} = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\Sigma m r}{\Sigma m}\right) \]
03

Apply the product rule for differentiation

We will use the product rule to differentiate the term inside the brackets: \[ \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\Sigma m r}{\Sigma m}\right) = \frac{(\Sigma m \cdot \dot{r} + r \cdot \dot{\Sigma m})}{(\Sigma m)^2} \]
04

Simplify the expression using the hint

Using the hint, we know that \(\Sigma m\) is constant and \(\Sigma \dot{m} r = 0\) between collisions. Thus, the differentiation term simplifies to: \[ \frac{\Sigma m \cdot \dot{r} + r \cdot \dot{\Sigma m}}{(\Sigma m)^2} = \frac{\Sigma m \cdot \dot{r}}{(\Sigma m)^2} \]
05

Determine the velocity of the center of mass

The velocity of the center of mass (\(\dot{\boldsymbol{r}}_{\mathrm{CM}}\)) can now be found from the simplified expression: \[ \dot{\boldsymbol{r}}_{\mathrm{CM}} = \frac{\Sigma m \cdot \dot{r}}{(\Sigma m)^2} \] Since, \(\Sigma m\) is constant, it can be factored out from both the numerator and denominator, which results in: \[ \dot{\boldsymbol{r}}_{\mathrm{CM}} = u_{\mathrm{CM}} \] This shows that the center of mass moves with the velocity of the center of mass frame.

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Most popular questions from this chapter

taking \(h=6.63 \times 10^{-27} \mathrm{crgs}\) and \(c=300000 \mathrm{~km} \mathrm{~s}^{-1}\), calculate how many photons of wavelength \(5 \times 10^{-5} \mathrm{~cm}\) (in the yellowgreen of the visible spectrum) must fall per second on a blackened plate to produce a force of one dyne. [Answer: \(7.3 \times 10^{21}\). Hint: force equals momentum absorbed per unit time.]

How fast must a particle move before its kinetic energy equals its rest energy? [0.866 c]

In an inertial frame there is a uniform electric field e in the positive \(x\)-direction, which exerts a rest mass preserving force \(\mathbf{f}=q \mathbf{e}\) on a particle carrying a charge \(q\). If the particle is released from rest at the origin, and \(m, \tau\) denote its rest mass and proper time, prove that the particle will move (hyperbolically) according to the equation

Prove that, in relativistic as in Newtonian mechanics, the time rate of change of the angular momentum \(h=\mathbf{r} \times p\) of a particle about an origin \(\mathrm{O}\) is equal to the couple \(\mathrm{r} \times \mathrm{f}\) of the applied force about \(\mathrm{O}\). If 1 \(^{\mu v}\) is the particle's four- angular momentum, and if we define the fourcouple \(G^{\mu v}=x^{\mu} F^{v}-x^{v} F^{\mu}\), where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), prove that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) and that the spacc-space part of this equation corresponds to the above threevector result.

15\. A rocket propels itself rectilinearly by emitting radiation in the direction opposite to its motion. If \(V\) is its final velocity relative to its initial rest frame, prove \(a b\) initio that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{1 / 2} $$ and compare this with the result of Exercise 5 above. [Hint: equate energies and momenta at the beginning and at the end of the acceleration, writing \(\Sigma h v\) and \(\Sigma h v / c\) for the total energy and momentum, respectively, of the emitted photons.]
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