/*! 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 71 The mass of a flywheel is \(5.6 ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 71

The mass of a flywheel is \(5.6 \times 10^{4} \mathrm{kg} .\) This particular flywheel has its mass concentrated at the rim of the wheel. If the radius of the wheel is \(2.6 \mathrm{m}\) and it is rotating at 350 rpm, what is the magnitude of its angular momentum?

Short Answer

Expert verified
Answer: The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.

Step by step solution

01

Determine the moment of inertia of the flywheel

Using the formula for a hoop's moment of inertia, calculate the moment of inertia (I) of the flywheel: I = MR^2 M = 5.6 * 10^4 kg (mass of the flywheel) R = 2.6 m (radius of the flywheel) I = (5.6 * 10^4 kg) * (2.6 m)^2 I = 3.7776 * 10^5 kg * m^2 (moment of inertia)
02

Convert rotational speed to angular velocity

The given rotational speed of the flywheel is 350 rpm. To convert rpm to radians per second, we can use the formula: 蠅 = (2 * 蟺 radians * rotational_speed) / 60 seconds rotational_speed = 350 rpm 蠅 = (2 * 蟺 * 350) / 60 蠅 鈮� 36.65 rad/s (angular velocity)
03

Calculate the angular momentum

Now that we have the moment of inertia and angular velocity, we can calculate the angular momentum (L) of the flywheel using the formula: L = I蠅 L = (3.7776 * 10^5 kg*m^2) * (36.65 rad/s) L 鈮� 1.3837 * 10^7 kg*m^2/s The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.

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Most popular questions from this chapter

A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of \(2.50 \mathrm{kg} \cdot \mathrm{m}^{2}\) when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to \(1.60 \mathrm{kg} \cdot \mathrm{m}^{2} ?\)
A mechanic turns a wrench using a force of \(25 \mathrm{N}\) at a distance of \(16 \mathrm{cm}\) from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench?
A 2.0 -kg uniform flat disk is thrown into the air with a linear speed of \(10.0 \mathrm{m} / \mathrm{s} .\) As it travels, the disk spins at 3.0 rev/s. If the radius of the disk is \(10.0 \mathrm{cm},\) what is the magnitude of its angular momentum?
Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$

A solid sphere is rolling without slipping or sliding down a board that is tilted at an angle of \(35^{\circ}\) with respect to the horizontal. What is its acceleration?
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