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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The concept of moment of inertia is pivotal in understanding rotational dynamics. Moment of inertia is essentially the rotational counterpart to mass in linear motion. It measures how challenging it is to change an object's rotation. The greater the moment of inertia, the harder it is to change its rotational state.

For a thin circular ring, the moment of inertia about its axis is calculated using the formula:
  • \[ I = M imes r^2 \]
Where \( M \) is the mass, and \( r \) is the radius. This formula indicates how the mass and distance from the rotation axis affect the moment of inertia. An increase in either the mass \( M \) or the radius \( r \) will increase \( I \).

Understanding its calculation helps in problems involving rotational dynamics, especially those centering on the conservation of angular momentum.
Angular Velocity
Angular velocity describes how quickly an object rotates or revolves around a point or an axis. It is typically denoted by the Greek letter \( \omega \) (omega). Think of it as the speed of rotation, akin to linear velocity in straight-line motion.

In rotational problems, angular velocity can change when the mass distribution around the axis of rotation changes, such as when extra masses are added to a rotating body. This change is particularly crucial when applying the conservation of angular momentum.

When masses are added to a rotating ring, the angular velocity changes to maintain constant angular momentum, assuming no external torques are applied. The formula for conservation of angular momentum will help us calculate the new angular velocity by equating the initial and final angular momenta.
Rotational Dynamics
Rotational dynamics deals with forces and torques and their impact on rotational motion. Just as Newton’s laws govern linear motion, similar principles govern rotational motion. These include concepts like torque, moment of inertia, and angular momentum.

In many physics problems like the one outlined here, understanding the link between these elements is crucial. For instance, when additional masses are added to the rotating ring, the system's moment of inertia changes, and consequently, to conserve angular momentum, the ring's angular velocity also changes.

This problem solved by applying angular momentum conservation shows a typical rotational dynamics scenario where no external torque is acting, so the system's angular momentum remains constant. This principle is widely used to solve various physics problems, allowing predictions of the rotational behavior of bodies after certain changes are made.
Physics Problem Solving
Solving physics problems often involves a strategic approach, particularly when dealing with concepts like rotational dynamics and conservation laws. Here's a breakdown of steps often taken:

  • Identify What is Given: List all known quantities such as mass, initial velocity, and radius.
  • Understand What is Needed: Know what you need to find, such as final velocity or momentum.
  • Apply Relevant Physical Principles: Use laws like conservation of angular momentum to set initial and final states equal when no outside influence (torque) is present.
  • Solve Algebraically: Often, before substituting numerical values, solve the equation algebraically for the desired quantity.
These steps ensure a clear path from understanding the problem to reaching a solution. Practice is key to mastering these techniques, and they serve as a backbone to tackling numerous physics scenarios.

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Most popular questions from this chapter

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\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

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 \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ...... \(\\{\mathrm{A}\\} \sqrt{2}\) \(\\{B\\} \sqrt{(1 / 2)}\) \(\\{\mathrm{C}\\}(1 / 2) \quad\\{\mathrm{D}\\} 2\)

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

An automobile engine develops \(100 \mathrm{kw}\) when rotating at a speed of 1800 r.p.m. what torque does it deliver? \(\\{\mathrm{A}\\} 350 \mathrm{Nm} \quad\\{\mathrm{B}\\} 440 \mathrm{Nm}\) \\{C \(\\} 531 \mathrm{Nm} \quad\\{\mathrm{D}\\} 628 \mathrm{Nm}\)

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