/*! 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 96 A doubling of the torque produce... [FREE SOLUTION] | 91Ó°ÊÓ

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A doubling of the torque produces a greater angular acceleration. Which of the following would do this, assuming that the tension in the rope doesn't change? (a) Increasing the pulley diameter by a factor of \(\sqrt{2} ;\) (b) increasing the pulley diameter by a factor of \(2 ;\) (c) increasing the pulley diameter by a factor of \(4 ;\) (d) decreasing the pulley diameter by a factor of \(\sqrt{2}\)

Short Answer

Expert verified
Increasing the pulley diameter by a factor of 2 (option b) will produce a greater angular acceleration by doubling the torque.

Step by step solution

01

Torque doubling solution

The formula for torque is \(Torque = rFsin(Ï´)\). Since the tension in the rope does not change, neither does the applied force. Therefore, F and \(sin(Ï´)\) are constant. The force and angle of application will not change regardless of the pulley's diameter. So, the formula can be simplified to \(Torque = rF\). The question now is how changing the radius (which is half the diameter) affects this equation.
02

Testing each option

We now need to check each option to see which one results in a doubling of the torque. (a) Increasing the pulley diameter by a factor of \(\sqrt{2}\) would result in the radius being increased by \(\sqrt{2}/2\), which does not double the original torque because \(rF(\sqrt{2}/2) \neq 2rF\). (b) Increasing the pulley diameter by a factor of 2 would result in the radius being doubled, which doubles the original torque because \(rF * 2 = 2rF\).(c) Increasing the pulley diameter by a factor of 4 would result in the radius being quadrupled, which doesn't double but quadruples the original torque because \(rF * 4 \neq 2rF\).(d) Decreasing the pulley diameter by a factor of \(\sqrt{2}\) would result in the radius being halved, which does not double, but halves the torque because \(rF / 2 \neq 2rF\).
03

Final conclusion

Using our understanding of how changes in radius affect torque, we are able to conclude that only by doubling the radius (which is achieved by doubling the pulley diameter) can we double the torque. Therefore, the only correct answer is option (b).

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Key Concepts

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

Torque Formula
Understanding the torque formula is essential when studying rotational dynamics. Torque (\tau) describes the rotational effect of a force applied at a distance from the pivot point. The mathematical representation of torque is given by ewlineewline \(Torque = rF\sin(\theta)\) ewlineewline where \(r\) is the radius, or the distance from the axis of rotation to the point where the force is applied, \(F\) is the magnitude of the force, and \(\theta\) is the angle between the force and the lever arm. For a rope wrapped around a pulley, as in the provided exercise, the angle is typically 90 degrees, which makes \(\sin(\theta)\) equal to 1. Thus, the formula simplifies to \(Torque = rF\).
  • Torque is directly proportional to both the force applied and the distance from the axis at which that force is exerted.
  • The larger the force or the longer the lever arm, the greater the torque produced.
ewlineewline This relationship is crucial to understand why changes in the pulley's diameter directly affect the torque produced, as seen in the exercise solution.
Radius Effect on Torque
The effect of the radius on torque is a pivotal concept when solving problems related to rotational motion. Since torque is the product of the radius and force, any changes in radius have a direct and proportional impact on torque. This is best illustrated with the formula: ewlineewline \(Torque = rF\). ewlineewline In the context of the problem presented, increasing the radius (or the effective lever arm) will accordingly increase the torque, provided that the force remains constant. As seen in the step-by-step solution:
  • An increase in diameter by a factor of 2 directly doubles the radius, which consequently doubles the torque.
  • A smaller or larger factor of change in the diameter will not result in the required doubling of torque.
ewlineewline It's important to note that this relationship between radius and torque can be applied universally, whether it's a simple pulley, a wrench turning a bolt, or gears in a machine. By manipulating the lever arm distance, the amount of torque can be finely controlled.
Angular Acceleration
Angular acceleration is another fundamental concept in the study of rotational motion. It is defined as the rate of change of angular velocity over time, similar to how linear acceleration is the rate of change of velocity.The formula for angular acceleration (\(\alpha\)) is: ewlineewline \(\alpha = \frac{\text{Δω}}{\text{Δt}}\) ewlineewline where \(\text{Δω}\) is the change in angular velocity and \(\text{Δt}\) is the change in time. Angular acceleration is directly related to torque through Newton's second law for rotation: ewlineewline \(\tau = I\alpha\) ewlineewline where \(\tau\) is the torque applied to a body, and \(I\) is the moment of inertia of the body.
  • The greater the torque applied to an object, the greater its angular acceleration, provided the moment of inertia remains constant.
  • For a given torque, a larger moment of inertia (which depends on both mass and radius) will result in a smaller angular acceleration.
ewlineewline This concept links back to the textbook exercise problem, which implies that by adjusting the torque, one can achieve a greater angular acceleration, assuming the same moment of inertia. By doubling the torque, as suggested by selecting the correct pulley diameter, one theoretically doubles the angular acceleration, allowing us to understand the interplay between these physical quantities in rotational dynamics.

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Most popular questions from this chapter

In your job as a mechanical engineer you are designing a flywheel and clutch- plate system like the one in Example \(10.11 .\) Disk \(A\) is made of a lighter material than disk \(B\), and the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B .\) The moment of inertia of the shaft is negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0} ; B\) is initially at rest. The accelerating torque is then removed from \(A,\) and \(A\) is coupled to \(B\). (Ignore bearing friction.) The design specifications allow for a maximum of \(2400 \mathrm{~J}\) of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk \(A\) so as not to exceed the maximum allowed value of the thermal energy?

A \(50.0 \mathrm{~kg}\) 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. \(\mathbf{P} 10.58\) ). The coefficient of kinetic friction 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 rev \(/ \min\) in \(9.00 \mathrm{~s} ?\) (b) After the grindstone attains an angular speed of 120 rev \(/ \mathrm{min}\), what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev \(/ \min ?\) (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 thin uniform rod has a length of \(0.500 \mathrm{~m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of \(0.400 \mathrm{rad} / \mathrm{s}\) and a moment of inertia about the axis of \(3.00 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is \(0.160 \mathrm{~m} / \mathrm{s}\). The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

A \(2.00 \mathrm{~kg}\) 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 \(0.800 \mathrm{~s}\) (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 metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

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