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A horizontal turn table in the form of a disc of radius \(r\) carries a gun at \(G\) and rotates with angular velocity \(\omega_{0}\) about a vertical axis passing through the centre \(O\). The increase in angular velocity of the system if the gun fires a bullet of mass \(m\) with a tangential velocity \(v\) with respect to the gun is (moment of inertia of gun \(+\) table about \(O\) is \(I_{0}\) ) (1) \(\frac{m v r}{I_{0}+m r^{2}}\) (2) \(\frac{2 m v r}{I_{0}}\) (3) \(\frac{v}{2 r}\) (4) \(\frac{m v r}{2 I_{0}}\)

Short Answer

Expert verified
The increase in angular velocity is \( \frac{m v r}{I_{0} + m r^2} \). (Option 1)

Step by step solution

01

Understand the Given Information

The problem involves a rotating turntable with a gun that fires a bullet. The turntable rotates with an initial angular velocity \( \omega_{0} \). The gun fires a bullet with a tangential velocity \( v \) and the bullet's mass is \( m \). The moment of inertia of the turntable and gun system is \( I_{0} \). We need to find the change in angular velocity after the bullet is fired.
02

Apply Conservation of Angular Momentum

Since no external torques act on the system, the angular momentum before and after firing the bullet is conserved. Before firing, the angular momentum is \( L_{i} = I_{0} \omega_{0} \). After firing, the angular momentum \( L_{f} \) of the system (turntable + bullet) is \( (I_{0} + m r^2) \omega \), where \( \omega \) is the new angular velocity.
03

Calculate the Angular Momentum of the Bullet

When the gun fires, the bullet gains an angular momentum \( L_{b} = mvr \) relative to the center \( O \) due to its tangential velocity \( v \). This bullet momentum must be considered in the conservation equation.
04

Write the Conservation Equation and Solve for \( \omega \)

The conservation of angular momentum equation is:\[ I_{0} \omega_{0} = (I_{0} + m r^2) \omega + m v r \]Rearrange to find the change in angular velocity \( \Delta \omega = \omega - \omega_{0} \):\[ \omega = \omega_{0} + \frac{m v r}{I_{0} + m r^2} \]
05

Identify the Increase in Angular Velocity

The increase in angular velocity \( \Delta \omega \) is \( \frac{m v r}{I_{0} + m r^2} \). This matches solution option (1).

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Key Concepts

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

Conservation of Angular Momentum
In rotational dynamics, the principle of conservation of angular momentum is a fundamental concept. It states that if no external torque acts on a system, the total angular momentum of the system remains constant. Angular momentum is the rotational equivalent of linear momentum and is represented as the product of moment of inertia and angular velocity.
In our exercise, a bullet is fired from a rotating turntable. Here, the total angular momentum before the bullet is fired is the product of the moment of inertia of the gun and table and its initial angular velocity. After the bullet is fired, the system's angular momentum includes both the rotational momentum of the bullet around the center of the turntable and the remaining momentum of the gun and table.
  • Before firing, angular momentum: \(L_i = I_{0} \omega_{0}\)
  • After firing, system's angular momentum: \(L_f = (I_{0} + mr^2) \omega\)
By setting these equal, as there are no external torques, we can solve for the change in the system's angular velocity. Understanding this concept is crucial for solving problems where internal forces change the distribution of mass and velocity within a system.
Moment of Inertia
The moment of inertia is an important quantity in rotational dynamics, describing how mass is distributed with respect to an axis of rotation. It acts as a rotational mass, essentially quantifying the resistance of an object to changes in its rotational state.
In the problem, the moment of inertia of the turntable and gun is given as \( I_{0} \). This value is crucial to determining the system's behavior when the bullet is fired. It's important to remember that for a turntable or disc, the moment of inertia not only depends on the mass of the system but also on how that mass is distributed concerning the rotation axis.
Additionally, the bullet contributes an extra term to the moment of inertia after it is fired, \( mr^2 \), because it acts as a point mass at a distance \( r \) from the axis of rotation. Appreciating this concept helps in understanding how different parts of a system contribute to its overall rotational dynamics:
  • Original inertia: \(I_{0}\)
  • Additional inertia due to bullet: \(mr^2\)
  • Effective inertia after firing: \(I_{0} + mr^2\)
This concept emphasizes how mass distribution can significantly affect rotational motion.
Angular Velocity
Angular velocity is a vector quantity that represents how fast an object rotates or revolves concerning an axis. It is measured in radians per second and indicates how fast the angular displacement occurs.
In this exercise, the system initially starts with an angular velocity of \( \omega_{0} \), and after the bullet is fired, its angular velocity changes. The increase in angular velocity is determined by the conservation of angular momentum. The additional velocity imparted by the bullet changes the overall momentum, prompting a new angular velocity for the system.
Computing the change in angular velocity involves understanding that the energy and momentum contributions of the bullet impact the system's dynamics. This is resolved through the relationship established by:
  • Initial angular velocity: \( \omega_0 \)
  • Final angular velocity: \( \omega = \omega_0 + \frac{mvr}{I_0 + m r^2} \)
  • Increase in angular velocity: \( \Delta \omega = \omega - \omega_0 = \frac{mvr}{I_0 + m r^2} \)
This showcases the effect of dynamic interactions within rotating bodies and how internal forces influence overall motion.

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