Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 620

A binary star consist of two stars \(\mathrm{A}(2.2 \mathrm{Ms})\) and \(\mathrm{B}\) (mass \(11 \mathrm{Ms}\) ) where \(\mathrm{Ms}\) is the mass of sun. They are separated by distance \(\mathrm{d}\) and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of \(\operatorname{star} B\). about the centre of mass is \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)

Short Answer

Expert verified
The short answer to the given question is \(\boxed{\text{(A)}\, 6}\).

Step by step solution

01

Find the center of mass

To find the center of mass, we can use the formula: \(\frac{m_A \cdot r_A + m_B \cdot r_B}{m_A + m_B}\), where \(m_A = 2.2 \,\mathrm{Ms}\) and \(m_B = 11 \,\mathrm{Ms}\) are the masses of the stars, and \(r_A\) and \(r_B\) are their respective distances from the center of mass. Since we only need the ratio of the distances, we can set up the equation: \(\frac{r_A}{r_B} = \frac{m_B}{m_A} = \frac{11 \,\mathrm{Ms}}{2.2 \,\mathrm{Ms}} = 5\). This gives us a ratio of distances, with star A being 5 times farther from the center of mass than star B.
02

Find the individual angular momenta

Next, let's find the angular momenta for star A and star B. Angular momentum (L) is given by the formula: \(L = m \cdot r \cdot v\), where m is the mass, r is the distance from the center of mass, and v is the tangential velocity. Since we only need the ratio of the angular momenta, and both stars have the same angular velocities, we can set up the equation: \(\frac{L_A}{L_B} = \frac{m_A \cdot r_A}{m_B \cdot r_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5}{11 \,\mathrm{Ms}}\).
03

Calculate the ratio of the total angular momentum to the angular momentum of star B

Now, we find the total angular momentum of the binary star system, which is the sum of the individual angular momenta: \(L_{total} = L_A + L_B\). To find the ratio of the total angular momentum to the angular momentum of star B, we use: \(\frac{L_{total}}{L_B} = \frac{L_A + L_B}{L_B}\). Substituting the values from Step 2, we get: \(\frac{L_{total}}{L_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5 + 11 \,\mathrm{Ms}}{11 \,\mathrm{Ms}}\). Solution: \(\frac{L_{total}}{L_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5 + 11 \,\mathrm{Ms}}{11 \,\mathrm{Ms}} = \boxed{1 + 5} = \boxed{6}\). The correct answer is \(\boxed{\text{(A)}\, 6}\).

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

A solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

A wheel rotates with a constant acceleration of $2.0 \mathrm{rad} / \mathrm{sec}^{2}$ If the wheel start from rest. The number of revolution it makes in the first ten seconds will be approximately. \(\\{\mathrm{A}\\} 8\) \\{B \\} 16 \(\\{\mathrm{C}\\} 24\) \(\\{\mathrm{D}\\} 32\)

Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio \(1: 3\). The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. \(\\{\mathrm{A}\\} 1: 3\) \\{B\\} \(3: 1\) \\{C\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)
See all solutions

Recommended explanations on English Textbooks

Pragmatics

Read Explanation

Single Paragraph Essay

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Phonetics

Read Explanation

5 Paragraph Essay

Read Explanation

Email

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free