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Chapter 5: Problem 573

A thin circular ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity \(\mathrm{w}\). Two objects each of mass \(\mathrm{m}\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity.... $\\{\mathrm{A}\\}[\\{\omega(\mathrm{M}-2 \mathrm{~m})\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]$ \(\\{\mathrm{B}\\}[\\{\omega \mathrm{M}\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]\) \(\\{C\\}[\\{\omega M)\\} /\\{M+m\\}]\) \(\\{\mathrm{D}\\}[\\{\omega(\mathrm{M}+2 \mathrm{~m})\\} / \mathrm{M}]\)

Short Answer

Expert verified
The new angular velocity of the rotating ring after two masses are gently attached to it is given by \( \omega' = \omega \frac{M}{M+2m} \).

Step by step solution

01

Define Initial and Final Angular Momentum

The initial angular momentum (L_initial) is the mass of the ring (M) times its initial angular velocity (ω). It can be represented as \( L_{initial} = I_{initial} * ω \), where \( I_{initial} \) is the moment of inertia of the ring which is \( I_{initial} = M*r^2 \). The final angular momentum (L_final) is the total moment of inertia (I_total) times the final angular velocity (ω'). It can be represented as \( L_{final} = I_{total} * ω' \), where \( I_{total} \) is the sum of the moment of inertia of the ring and the two added masses. It can be calculated as \( I_{total} = I_{initial} + 2*m*r^2 \).
02

Apply the Conservation of Angular Momentum

According to the conservation of angular momentum, the initial and final angular momentum of the system must be equal if no external torque is applied. So, we set \( L_{initial} = L_{final} \). Substituting our values, we get \[ I_{initial} * ω = I_{total} * ω' \].
03

Solve for the Final Angular Velocity

Solving for ω', we get \[ ω' = \frac{I_{initial} * ω}{I_{total}} \]. Substitute the values of \( I_{initial} \) and \( I_{total} \) into the equation, then find out \( ω' \).
04

Match the Final Angular Velocity with the Given Choices

Finally, you should compare this final angular velocity ω' to the given choices and find out which of these matches to your computed final angular velocity. Without explicitly given numerical values for m, M, r, and ω, we cannot determine a precise numerical answer but the procedure contains all the algebraic manipulations needed if values would be given. The correct answer with respect to the form that it is presented in the options is \( \omega \frac{M}{M+2m} \).

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Most popular questions from this chapter

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

A particle performing uniform circular motion has angular momentum \(L\)., its angular frequency is doubled and its \(K . E\). halved, then the new angular momentum is \(\\{\mathrm{A}\\} 1 / 2\) \\{B \(\\} 1 / 4\) \(\\{\mathrm{C}\\} 2 \mathrm{~L}\) \(\\{\mathrm{D}\\} 4 \mathrm{~L}\)

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

A cylinder of mass \(5 \mathrm{~kg}\) and radius \(30 \mathrm{~cm}\), and free to rotate about its axis, receives an angular impulse of $3 \mathrm{~kg} \mathrm{M}^{2} \mathrm{~S}^{-1}$ initially followed by a similar impulse after every \(4 \mathrm{sec}\). what is the angular speed of the cylinder 30 sec after initial impulse? The cylinder is at rest initially. \(\\{\mathrm{A}\\} 106.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{B\\} \(206.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{C\\} \(107.6 \mathrm{rad} \mathrm{S}^{-1}\) \\{D \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)
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