/*! 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

A small \(10.0 \mathrm{~g}\) 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?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

A grindstone in the shape of a solid disk with diameter \(0.520 \mathrm{~m}\) and a mass of \(50.0 \mathrm{~kg}\) is rotating at \(850 \mathrm{rev} / \mathrm{min} .\) You press an ax against the rim with a normal force of \(160 \mathrm{~N}\) (Fig. \(\mathrm{P} 10.58\) ), and the grindstone comes to rest in \(7.50 \mathrm{~s}\). Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

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