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Chapter 8: Problem 50

A constant torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) is applied to a grindstone whose moment of inertia is \(0.130 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Using energy principles and neglecting friction, find the angular speed after the grindstone has made \(15.0\) revolutions, Hint: The angular equivalent of \(W_{\mathrm{net}}=F \Delta x=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) is \(W_{\text {net }}\) \(=\tau \Delta \theta=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} J \omega_{i}^{2} .\) You should convince yourself that this relationship is correct.

Short Answer

Expert verified
The final angular speed after the grindstone has made 15.0 revolutions under a constant torque of 25.0 Nm is found by applying the given formula. This results in \(\omega_{f} = \sqrt{\frac{ 2 \times 25.0 \mathrm{~N} \cdot \mathrm{m} \times 30\pi \mathrm{~rad} }{ 0.130 \mathrm{~kg} \cdot \mathrm{m}^{2} } }\) rad/s

Step by step solution

01

Understanding the given information

To start, note all the important values given: the applied torque, \( \tau = 25.0 \mathrm{~N} \cdot \mathrm{m} \), the grindstone's moment of inertia, \( I = 0.130 \mathrm{~kg} \cdot \mathrm{m}^{2} \), and the number of revolutions, which is 15. We also know that the initial angular speed \( \omega_{i} \) is 0, since the grindstone starts from rest.
02

Converting revolutions into radians

Recall that one revolution is equivalent to \( 2\pi \) radians. Therefore, 15 revolutions would be \( 15 \times 2\pi = 30\pi \) radians.
03

Applying the formula

Put the given values into the formula provided as the hint: \( W_{\text {net }} = \tau \Delta \theta = \frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} I \omega_{i}^{2} \). Therefore, it can be written as \( 25.0 \mathrm{~N} \cdot \mathrm{m} \times 30\pi \mathrm{~rad} = \frac{1}{2} \times 0.130 \mathrm{~kg} \cdot \mathrm{m}^{2} \times \omega_{f}^{2} - \frac{1}{2} \times 0.130 \mathrm{~kg} \cdot \mathrm{m}^{2} \times 0^{2} \).
04

Solving for the final angular speed

Simplifying the above equation, we can find the final angular speed, \( \omega_{f} = \sqrt{\frac{ 2 \times 25.0 \mathrm{~N} \cdot \mathrm{m} \times 30\pi \mathrm{~rad} }{ 0.130 \mathrm{~kg} \cdot \mathrm{m}^{2} } } \).

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Key Concepts

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

Torque
Torque is the rotational equivalent of force. Imagine trying to open a door: it's easier if you apply force at the handle rather than closer to the hinge. This is because torque (\( \tau \)) depends not just on the amount of force, but also on how far that force is applied from a pivot point (radius). Torque is calculated as:
  • \( \tau = r \times F \)
  • where \( r \) is the radius, and \( F \) is the force applied.
In our exercise, the torque of \(25.0 \, \mathrm{N} \cdot \mathrm{m}\) represents how the force is making the grindstone spin. Just like in linear motion, torque can change an object's rotational state, speeding it up, slowing it down, or keeping it moving at a constant speed.
Moment of Inertia
The moment of inertia (\( I \)) is like the rotational version of mass. It tells us how much resistance an object has to changes in its rotational motion. Consider how spinning a compact disk (CD) requires different effort compared to a bicycle wheel. This is reflected in their different moments of inertia.The moment of inertia depends on:
  • Mass distribution - how that mass is spread out relative to the axis of rotation.
  • More mass far from the axis results in a larger \( I \).
In the problem, the grindstone has a moment of inertia of \(0.130 \, \mathrm{kg} \cdot \mathrm{m}^2\), a value that affects how easily it can be set into rotational motion by the torque applied.
Energy Principles
Energy principles involve understanding how energy transforms and transfers within a system. In rotational motion, we use them to find how work done by torque affects an object's motion. The principle used here is akin to the work-energy principle in linear motion. It is represented as:
  • \( W_{\text {net }} = \tau \Delta \theta \)
  • It is the rotational work done, equating to changes in rotational kinetic energy.
  • The formula used \( \frac{1}{2} I \omega_{f}^{2} - \frac{1}{2} I \omega_{i}^{2} \) relates the initial and final rotational states.
For the grindstone, work done by the torque turns into kinetic energy, resulting in a final rotational speed. By applying the given torque over the 15 revolutions, its energy increases, bringing a rise in its motion.
Angular Speed
Angular speed (\( \omega \)) measures how fast something rotates. It's akin to linear speed but for rotation. Angular speed tells you how many radians an object covers per second. The relationship is straightforward:
  • \( \omega = \frac{\theta}{t} \)
  • \( \theta \) being the angular displacement in radians.
  • More simply, \( \omega \) can be affected by changes in torque or moment of inertia.
In the exercise, we calculate the final angular speed (\( \omega_f \)) once we have determined how the applied torque over multiple revolutions has changed the grindstone's motion. By understanding this process, we can see how the rotational motion evolves, transforming initial energy input into the grindstone's rapid spinning action.

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Most popular questions from this chapter

]A \(150-\mathrm{kg}\) merry-go-round in the shape of a uniform, solid, horizontal disk of radius \(1.50 \mathrm{~m}\) is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of \(0.500\) rev/s in \(2.00 \mathrm{~s}\) ?

A car is designed to get its energy from a rotating flywheel with a radius of \(2.00 \mathrm{~m}\) and a mass of \(500 \mathrm{~kg}\). Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to \(5000 \mathrm{rev} / \mathrm{min}\). (a) Find the kinetic energy stored in the flywheel. (b) If the flywheel is to supply energy to the car as a \(10.0\) -hp motor would, find the length of time the car could run before the flywheel would have to be brought back up to speed.

Two astronauts (Fig. \(\mathrm{P} 8.72\) ), each having a mass of \(75.0 \mathrm{~kg}\), are connected by a \(10.0-\mathrm{m}\) rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(5.00 \mathrm{~m}\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

A model airplane with mass \(0.750 \mathrm{~kg}\) is tethered by a wire so that it flies in a circle \(30.0 \mathrm{~m}\) in radius. The airplane engine provides a net thrust of \(0.800 \mathrm{~N}\) perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to jts flight path.

An Atwood's machine consists of blocks of masses \(m_{1}=10.0 \mathrm{~kg}\) and \(m_{2}=20.0 \mathrm{~kg}\) attached by a cord running over a pulley as in Figure \(\mathrm{P} 8.40 .\) The pulley is a solid cylinder with mass \(M=8.00 \mathrm{~kg}\) and radius \(r=0.200 \mathrm{~m}\) The block of mass \(m_{2}\) is allowed to drop, and the cord turns the pulley without slipping. (a) Why must the tension \(T_{2}\) be greater than the tension \(T_{1} ?\) (b) What is the acceleration of the system, assuming the pulley axis is frictionless? (c) Find the tensions \(T_{1}\) and \(T_{2}\).
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