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A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

Short Answer

Expert verified
(a) Angular speed is approximately 0.12 rad/s. (b) Total kinetic energy is 0.00032 J. (c) Energy comes from the bug's stored chemical energy.

Step by step solution

01

Understand the Conservation of Angular Momentum

Since the system is isolated and there are no external torques, angular momentum is conserved. Initially, the bar and bug have zero angular momentum since they are at rest. After the bug jumps, we need to account for their respective angular momentums. The angular momentum before the jump is equal to the angular momentum after the jump.
02

Calculate the Initial and Final Angular Momenta

Initial angular momentum (L_initial) is zero because the system is at rest. \( L_{initial} = 0 \).After the bug jumps, we have:- Bug’s contribution: \( L_{bug} = m_b \, v_b \, r \), where - \( m_b = 0.01 \, \mathrm{kg} \) is the bug’s mass, - \( v_b = 0.20 \, \mathrm{m/s} \) is the speed, - \( r = 1.0 \, \mathrm{m} \) is the initial position of the bug.- Bar’s contribution: \( L_{bar} = I_{bar} \omega \), where - \( I_{bar} = \frac{1}{3} M L^2 \) is the moment of inertia of the bar, - \( M = 0.05 \, \mathrm{kg} \), - \( L = 1.0 \, \mathrm{m} \), - \( \omega \) is the angular speed.Using conservation: \[ L_{initial} = L_{bug} - L_{bar} \Rightarrow 0 = m_b \, v_b \, r - I_{bar} \omega \]
03

Solve for Angular Speed \( \omega \)

Using the equation from the conservation of angular momentum:\[ 0 = m_b \, v_b \, r - \frac{1}{3} M L^2 \omega \]Substitute known values:\[ 0 = (0.01)(0.20)(1.0) - \frac{1}{3}(0.05)(1.0)^2 \omega \]\[ \Rightarrow 0 = 0.002 - \frac{1}{3}(0.05)(1) \omega \]\[ \Rightarrow 0 = 0.002 - 0.0167 \omega \]\[ \Rightarrow 0.0167 \omega = 0.002 \]\[ \Rightarrow \omega \approx 0.12 \, \mathrm{rad/s} \]The angular speed of the bar is approximately \( 0.12 \, \mathrm{rad/s} \).
04

Find the Total Kinetic Energy After the Bug Jumps

Calculate the kinetic energy of both the bug and the bar.- Bug’s kinetic energy: \[ KE_{bug} = \frac{1}{2} m_b v_b^2 = \frac{1}{2}(0.01)(0.20)^2 = 0.0002 \, \mathrm{J} \]- Bar’s rotational kinetic energy: \[ KE_{bar} = \frac{1}{2} I_{bar} \omega^2 = \frac{1}{2} \left( \frac{1}{3}(0.05)(1.0)^2 \right) (0.12)^2 \] \[ KE_{bar} \approx \frac{1}{2} (0.0167) (0.0144) = 0.00012 \, \mathrm{J} \]Total kinetic energy:\[ KE_{total} = KE_{bug} + KE_{bar} \approx 0.0002 + 0.00012 = 0.00032 \, \mathrm{J} \]
05

Source of the Kinetic Energy

The kinetic energy came from the chemical energy within the bug. As the bug jumps, it uses energy stored within its muscles to perform the leap, converting stored energy into kinetic energy.

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Key Concepts

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

Angular Speed
In the world of physics, angular speed is a measure of how quickly an object rotates around a particular axis. It's basically the speed at which the angle of rotation changes. For instance, when the bug jumps off the bar, it causes the bar to rotate about the nail. The angular speed of the bar is the rate at which this rotation occurs.
Angular speed is represented by the symbol \( \omega \) (omega) and is measured in radians per second (rad/s). One important thing to remember is that angular speed is different from linear speed. While linear speed refers to the rate at which an object moves along a path, angular speed deals with how fast an object spins.
In our example, once the bug jumps off, it contributes to the angular momentum of the system, enabling us to determine the angular speed \( \omega \) of the bar using the principle of conservation of angular momentum.
Moment of Inertia
Moment of inertia is a key concept in rotational dynamics. It is essentially a measure of an object's resistance to changes in its rotational motion and depends on the mass and shape of the object as well as how the mass is distributed relative to the axis of rotation.
  • In our context, the moment of inertia of the bar changes how it reacts when the bug leaps off.
  • Moment of inertia is symbolized by \( I \) and is typically expressed in units of kg·m².
For a thin bar pivoting about its end, the moment of inertia \( I_{bar} \) is given by \( \frac{1}{3} ML^2 \). Here, \( M \) represents the mass of the bar and \( L \) its length.
This formula indicates that the further away the mass is from the pivot, the larger the moment of inertia will be, making it more resistant to rotational acceleration. This means it requires more effort to change its state of motion. Thus, knowing the moment of inertia allows us to predict how much the bar will begin to rotate when influenced by the bug's leap.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. After the bug jumps from the bar, both the bug and the bar have kinetic energy. Whether the motion is translational, rotational, or a combination of both, kinetic energy is present.
In this problem, the bug exhibits translational kinetic energy, calculated with the formula \( KE_{bug} = \frac{1}{2} m_b v_b^2 \). On the other hand, the bar's kinetic energy is rotational. It's found by the equation \( KE_{bar} = \frac{1}{2} I_{bar} \omega^2 \). Both types of kinetic energy are measured in joules (J).
  • Translational kinetic energy is associated with the linear movement of the bug.
  • Rotational kinetic energy is tied to the spinning of the bar.
By adding these two kinetic energies, we find the total kinetic energy of the system. The source of all the kinetic energy here is primarily from the chemical energy stored in the bug, which it converts during its leap.
Physics Problem Solving
Solving physics problems often requires understanding the principles and formulas involved, breaking down the problem step-by-step, and ensuring all relevant factors are considered. A methodical approach can bring clarity to complex problems.
Let's break it down:
  • Start by understanding the main concept, like the conservation of angular momentum in this case. Identify what's conserved and why. Here, due to the lack of external torque, angular momentum is conserved.
  • Calculate the known quantities first, such as initial conditions like momentum or energy, to find the unknowns.
  • Apply relevant formulas, such as those for angular speed or kinetic energy, to get to the solution.
Sound physics problem-solving skills also involve verifying the solution to ensure it is logically consistent and practically possible, offering a deeper comprehension of the subject. Remember, practice enhancing these skills helps in tackling a wide range of physics problems.

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