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A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

Short Answer

Expert verified
(a) The angular speed is 0.645Ï€ rad/s when the rings reach the ends. (b) After the rings fly off, the angular speed is 3.225Ï€ rad/s.

Step by step solution

01

Understand Initial Conditions

The initial moment of inertia can be calculated considering both the rod and the rings. The rod rotates about its center: \[ I_{rod} = \frac{1}{12}mL^2 = \frac{1}{12}(0.0300 \, \text{kg})(0.400 \, \text{m})^2 = 4.0 \times 10^{-4} \, \text{kg} \, \text{m}^2 \]Each ring initially is at 0.0500 m from the center: \[ I_{rings}^{initial} = 2 \times (0.0200 \, \text{kg})(0.0500 \, \text{m})^2 = 1.0 \times 10^{-5} \, \text{kg} \, \text{m}^2 \]Total initial moment of inertia is \[ I_{initial} = I_{rod} + I_{rings}^{initial} = 4.0 \times 10^{-4} + 1.0 \times 10^{-5} = 4.1 \times 10^{-4} \, \text{kg} \, \text{m}^2 \]
02

Calculate Initial Angular Velocity

The system's initial angular velocity \( \omega_i \) in radians per second is calculated by converting revolutions per minute (rpm) to radians per second:\[ \omega_i = 30.0 \, \text{rev/min} \times \frac{2 \pi \, \text{rad}}{1 \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \pi \, \text{rad/s} \]
03

Calculate Moment of Inertia When Rings Reach the Ends

When the rings reach the ends, they are 0.200 m from the center:\[ I_{rings}^{final} = 2 \times (0.0200 \, \text{kg})(0.200 \, \text{m})^2 = 1.6 \times 10^{-3} \, \text{kg} \, \text{m}^2 \]The total moment of inertia then is: \[ I_{final} = I_{rod} + I_{rings}^{final} = 4.0 \times 10^{-4} + 1.6 \times 10^{-3} = 2.0 \times 10^{-3} \, \text{kg} \, \text{m}^2 \]
04

Use Conservation of Angular Momentum Until Rings Slide Off

Using the principle of conservation of angular momentum, the moment of inertia and angular velocity relationship is:\[ I_{initial} \omega_i = I_{final} \omega_f \]Plugging in the values:\[ (4.1 \times 10^{-4} \, \text{kg} \, \text{m}^2)(\pi \, \text{rad/s}) = (2.0 \times 10^{-3} \, \text{kg} \, \text{m}^2)\omega_f \]Solve for \( \omega_f \):\[ \omega_f = \frac{(4.1 \times 10^{-4})(\pi)}{2.0 \times 10^{-3}} = 0.645 \pi \, \text{rad/s} \]
05

Determine Angular Speed After Rings Fly Off

After the rings leave, only the rod contributes to the moment of inertia, which remains:\[ I_{rod} = 4.0 \times 10^{-4} \, \text{kg} \, \text{m}^2 \]Again, using conservation of angular momentum:\[ I_{final} \omega_f = I_{rod} \omega_{final} \]Plugging in known values:\[ (2.0 \times 10^{-3} \, \text{kg} \, \text{m}^2)(0.645 \pi \, \text{rad/s}) = (4.0 \times 10^{-4} \, \text{kg} \, \text{m}^2)\omega_{final} \]Solve for \( \omega_{final} \):\[ \omega_{final} = \frac{(2.0 \times 10^{-3} \times 0.645 \pi)}{4.0 \times 10^{-4}} = 3.225 \pi \, \text{rad/s} \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a measure of how much an object resists changes to its rotation. It is essentially the rotational equivalent of mass in linear dynamics. Here, the concept is crucial for understanding how the rotation of the system changes depending on the position of the mass along the rod.
The formula for the moment of inertia depends on the mass distribution relative to the axis of rotation. For a rod rotating about an axis through its center, the moment of inertia, denoted as \( I_{rod} \), is given by:\[I_{rod} = \frac{1}{12} m L^2\]where \( m \) is the mass of the rod and \( L \) is its length.
  • When the two rings are close to the center, their individual contribution to inertia is smaller compared to when they are at the ends. This changes because moment of inertia increases with the square of the distance from the axis.
  • As the rings move outward, the system's moment of inertia increases since the rings' mass distribution changes relative to the center.
This proportionality to the distance squared is why the calculation changes when the position of the rings changes along the rod.
Angular Velocity
Angular velocity describes how fast an object is rotating. Unlike linear velocity, which measures motion along a straight path, angular velocity measures how quickly something spins around an axis.
The initial angular velocity of the system in this exercise is given in revolutions per minute (rpm) and needs to be converted to radians per second (rad/s) for calculations.\[\omega_i = 30.0 \text{ rev/min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \pi \text{ rad/s}\]Angular velocity is crucial when applying the conservation of angular momentum. It helps determine how changes in moment of inertia affect rotational speed:
  • Initially, the angular velocity is higher because the moment of inertia is lower when the rings are near the center.
  • As rings move outwards, they increase the system's moment of inertia, causing the angular velocity to decrease if no external torques act on the system.
This relationship shows the inverse proportional effect between moment of inertia and angular velocity within the rotation.
Rotational Dynamics
Rotational dynamics involves the motion of objects that rotate. Unlike linear dynamics, where forces and motion happen in a straight line, rotational dynamics deals with torques, moments of inertia, and angular momentum.
The core principle used here is the conservation of angular momentum. This principle states that if no external torques act on a system, the total angular momentum remains constant.
Angular momentum \( L \) is defined as the product of moment of inertia \( I \) and angular velocity \( \omega \):\[L = I \cdot \omega\]In this exercise, the conservation principle is applied twice:
  • Firstly, when the rings are sliding outwards, their motion to the end affects the system's moment of inertia and thus must affect its rotational speed as per\[I_{initial} \omega_i = I_{final} \omega_f\]where the initial and final symbols denote states before and after the rings reach the ends of the rod.
  • Secondly, once the rings leave the rod, the rod's inertia alone dictates the new angular speed, requiring a fresh application of the conservation law.
    This dynamic shift demonstrates how interconnected moment of inertia and angular velocity are in establishing rotational behavior.

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Most popular questions from this chapter

A \(55-\mathrm{kg}\) runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 \(\mathrm{m} / \mathrm{s}\) . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 \(\mathrm{rad} / \mathrm{s}\) relative to the earth. The radius of the turntable is \(3.0 \mathrm{m},\) and its moment of inertia about the axis of rotation is 80 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

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In a physics laboratory you do the following ballistic pen- dulum experiment: You shoot a ball of mass \(m\) horizontally from a spring gun with a speed \(v\) . The ball is immediately caught a distance \(r\) below a frictionless pivot by a pivoted catcher assembly of mass \(M\) . The moment of inertia of this assembly about its rotation axis through the pivot is \(L\) . The distance \(r\) is much greater than the radius of the ball. (a) Use conservation of angular momentum to show that the angular speed of the ball and catcher just after the ball is caught is \(\omega=\operatorname{mor} /\left(m r^{2}+I\right) .\) (b) After the ball is caught, the center of mass of the ball-catcher assembly system swings up with a maximum height increase \(h\) . Use conservation of energy to show that \(\omega=\sqrt{2(M+m) g h /\left(m r^{2}+1\right)} .\) (c) Your lab partner says that linear momentum is conserved in the collision and derives the expression \(m v=(m+M) V,\) where \(V\) is the speed of the ball immediately after the collision. She then uses conservation of energy to derive that \(V=\sqrt{2 g h},\) so that \(m v=(m+M) \sqrt{2 g h} .\) Use the results of parts (a) and (b) to show that this equation is satisfied only for the special case when \(r\) is given by \(I=M r^{2} .\)

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