Let \(\overrightarrow{\mathbf{r}}_{\mathrm{cm}}\) be the position vector of the center of mass \(C\) of a system of particles with respect to the origin \(O\) of an inertial reference frame, and let \(\overrightarrow{\mathbf{r}}_{i}^{\prime}\) be the position vector of the \(i\) th particle, of mass \(m_{i}\), with respect to the center of mass \(C\). Hence \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\) (see Fig. \(\left.10-22\right)\). Now define the total angular momentum of the system of particles relative to the center of mass \(C\) to be \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right), \quad\) where \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i}^{\prime} / d t\) (a) Show that \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i} / d t-\) \(m_{i} d \overrightarrow{\mathbf{r}}_{\mathrm{cm}} / d t=\overrightarrow{\mathbf{p}}_{i}-m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}\) (b) Show next that \(d \overrightarrow{\mathbf{L}}^{\prime} / d t=\) \(\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times d \overrightarrow{\mathbf{p}}_{i}^{\prime} / d t\right) .(c)\) Combine the results of \((a)\) and \((b)\) and, using the definition of center of mass and Newton's third law, show that \(\vec{\tau}_{\mathrm{ext}}^{\prime}=d \overrightarrow{\mathbf{L}}^{\prime} / d t\), where \(\vec{\tau}_{\mathrm{ext}}^{\prime}\) is the sum of all the \(\mathrm{ex}\) - ternal torques acting on the system about its center of mass.

Step by step solution

01

Calculating momentum of ith particle

Given the position vector of the ith particle with respect to its center mass is \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\), we differentiate this with respect to time on both sides to get:\(d\overrightarrow{\mathbf{r}}_i/dt = d\overrightarrow{\mathbf{r}}_{cm}/dt + d\overrightarrow{\mathbf{r}}_{i}^{\prime}/dt\)\nThe momentum of the ith particle is given by \(m_i d\overrightarrow{\mathbf{r}}_i/dt = \overrightarrow{\mathbf{p}}_i\). Similarly, \(m_i d\overrightarrow{\mathbf{r}}_{cm}/dt\) is the momentum of the center of mass. Hence, by substituting from the previously obtained expression, we obtain \(m_i d\overrightarrow{\mathbf{r}}_i^{\prime}/dt = \overrightarrow{\mathbf{p}}_i - m_i d\overrightarrow{\mathbf{r}}_{cm}/dt = \overrightarrow{\mathbf{p}}_i - m_i \overrightarrow{\mathbf{v}}_{cm}\). Hence proved.

02

Deriving the time-rate change of angular momentum of the system

By definition, the total angular momentum is given by \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime}\times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right)\). Differentiating this with respect to time, we get:\(d\overrightarrow{\mathbf{L}}^{\prime}/dt = \Sigma(d(\overrightarrow{\mathbf{r}}_{i}^{\prime}\times \overrightarrow{\mathbf{p}}_{i}^{\prime})/dt)\). Using the product rule for differentiation, we get: \(d\overrightarrow{\mathbf{L}}^{\prime}/dt = \Sigma(\overrightarrow{\mathbf{r}}_{i}^{\prime}\times d\overrightarrow{\mathbf{p}}_{i}^{\prime}/dt + d\overrightarrow{\mathbf{r}}_{i}^{\prime}/dt\times\overrightarrow{\mathbf{p}}_{i}^{\prime})\)

03

Proving the relation with external torques

Using the results of above steps and the definition of center of mass and Newton's third law, we can show that the sum of external torques acting on the system equals the rate of change of angular momentum. The external torque exerted on the system is \(\vec{\tau}_{\mathrm{ext}}^{\prime} = d\overrightarrow{\mathbf{L}}^{\prime}/dt\). Here, by plugging the results from Step 2, we get the required proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!