/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A small 10.0-g bug stands at one... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 47

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: 0.48 rad/s (b) Total kinetic energy: 0.00212 J (c) Energy comes from the bug's chemical energy converted into motion.

Step by step solution

01

Understand the Concepts

The problem involves a bug jumping off a rotating bar, which can be analyzed using the concepts of angular momentum, conservation of angular momentum, and kinetic energy. The mass m of the bug is 10.0 g, the mass M of the bar is 50.0 g, and the length L of the bar is 100 cm. The bug jumps off with a speed v of 20.0 cm/s relative to the table.
02

Calculate Initial Angular Momentum

Because angular momentum is conserved, we need to find the initial angular momentum of the system. Initially, the bar does not rotate, so only the bug's motion contributes. The initial angular momentum L_initial is given by: \[ L_{initial} = m imes v imes L \] where m = 10 \, \text{g} = 0.01 \, \text{kg}, v = 20 \, \text{cm/s} = 0.2 \, \text{m/s}, and L = 0.5 \, \text{m}. Calculating gives: \[ L_{initial} = 0.01 \, \text{kg} \times 0.2 \, \text{m/s} \times 0.5 \, \text{m} = 0.001 \, \text{kg} \cdot \text{m}^2/\text{s} \].
03

Apply Conservation of Angular Momentum

The system's angular momentum remains constant, so the final angular momentum L_final (just after the bug jumps) of the bar equals the initial angular momentum of the system. The final angular momentum is given by: \[ L_{final} = I_{bar} \times \omega \] where \( I_{bar} = \frac{1}{3} M L^2 \) is the moment of inertia of the bar about the pivot point and \( \omega \) is the angular speed of the bar.
04

Solve for Angular Speed

We equate the initial and final angular momentum to solve for \( \omega \): \[ L_{initial} = \frac{1}{3} M L^2 \times \omega \]. Substitute \( M = 0.05 \, \text{kg} \), \( L = 0.5 \, \text{m} \): \[ 0.001 = \frac{1}{3} \times 0.05 \times (0.5)^2 \times \omega \]. Solving for \( \omega \): \[ \omega = \frac{0.001}{(\frac{1}{3} \times 0.05 \times 0.25)} = 0.48 \, \text{rad/s} \].
05

Calculate Total Kinetic Energy

The total kinetic energy just after the leap includes the rotational kinetic energy of the bar and the translational kinetic energy of the bug. The rotational kinetic energy (KE_rotational) of the bar is: \[ KE_{rotational} = \frac{1}{2} I_{bar} \omega^2 \]. The translational kinetic energy (KE_translational) of the bug is: \[ KE_{translational} = \frac{1}{2} m v^2 \]. Calculate each part using previous values: \[ KE_{rotational} = \frac{1}{2} \times \frac{1}{3} \times 0.05 \times (0.5)^2 \times (0.48)^2 = 0.00192 \, \text{J} \], \[ KE_{translational} = \frac{1}{2} \times 0.01 \times (0.2)^2 = 0.0002 \, \text{J} \]. Add both to find the total: \[ KE_{total} = KE_{rotational} + KE_{translational} = 0.00192 + 0.0002 = 0.00212 \, \text{J} \].
06

Understand Energy Source

The energy in the system after the bug leaps comes from the work done by the bug to jump off the bar. The chemical energy stored in the bug is converted into kinetic energy as it leaps and the bar rotates.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Any object that moves possesses kinetic energy. In this exercise, both the bug and the bar have kinetic energy after the bug jumps. If you imagine a moving car, it has kinetic energy because it is in motion. Similarly, when the bug leaps off the bar at a speed of 20 cm/s, it acquires kinetic energy.
The kinetic energy of the bug can be calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity. We replace these with the bug's mass (0.01 kg) and its velocity (0.2 m/s) to find its kinetic energy.
  • Translational kinetic energy: Energy from moving in a straight line.
  • Rotational kinetic energy: Energy from rotating around an axis.
In our problem, both types are involved. The bug has translational kinetic energy as it moves away from the bar. The bar itself has rotational kinetic energy because it spins around the pivot.
Moment of Inertia
The moment of inertia, often denoted by \( I \), is a measure of an object's resistance to changing its rotational motion. It's analogous to mass in linear motion. The more "spread out" the mass is from the axis of rotation, the larger the moment of inertia, making the object harder to spin.
In this instance, the thin bar rotating around its pivot has a moment of inertia calculated using the formula \( I = \frac{1}{3} ML^2 \), where \( M \) is the mass and \( L \) is the length of the bar. This is because it's a uniform bar rotating about one of its ends.
  • Higher moment of inertia: More resistance to changes in rotation.
  • The bar's shape affects \( I \): Uniform bars have specific formulas based on the axis.
Understanding the moment of inertia helps us see why heavier or longer objects rotate differently. The bar's moment of inertia, combined with the jump, allows us to compute its angular speed just after the bug departs.
Rotational Motion
Rotational motion occurs when an object spins around a fixed point or axis. In the bug-bar exercise, the bar鈥檚 rotational motion is triggered by the conservation of angular momentum when the bug jumps off.
Angular momentum, crucial in such problems, is conserved in the absence of an external torque. The initial angular momentum, from the bug's jump, becomes the bar's angular momentum after the leap. This concept ensures that we can calculate the bar's angular velocity \( \omega \) using \( L = I \omega \). By knowing the initial conditions (the bug's motion) and the system's inertial properties (moment of inertia), we can deduce the motion of the bar.
  • Angular momentum conservation: What goes in, comes out, due to no external torque.
  • Rotational equations help us model how an object behaves while spinning.
Understanding these principles makes it easier to solve problems, like our exercise, where rotational motion is key to analyzing movement after external interactions.

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Most popular questions from this chapter

A child rolls a 0.600 -kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s} .\) When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

A \(50.0\)-kg grindstone is a solid disk 0.520 \(\mathrm{m}\) in diameter. You press an ax down on the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57). The coefficient of kinetion between the blade and the stone is 0.60 , and there is a constant friction torque of 6.50 \(\mathrm{N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle 0.500 \(\mathrm{m}\) long to bring the stone from rest to 120 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (c) How much time does it take the grindstone to come from 120 \(\mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

A 2.00 -kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000 \(\mathrm{rev} / \mathrm{min}\) . (b) A drum with negligible mass, 0.400 \(\mathrm{m}\) in diameter, is attached to the motor shaft, and the power output of the motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the motor lift at constant speed? (c) At what constant speed will the weight rise?

A 42.0 -cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of 25.0 \(\mathrm{g} / \mathrm{cm} .\) This wheel is released from rest at the top of a hill 58.0 \(\mathrm{m}\) high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?
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