/*! 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 solid ball is released from rest and slides down a hillside that slopes downward at \(65.0^{\circ}\) from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2 . The kinetic energy of ball 2 is \(27.0 \mathrm{~J}\). What is the kinetic energy of ball \(1 ?\)

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?

A solid cylinder with radius \(0.140 \mathrm{~m}\) is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at \(t=0\) a constant horizontal force of \(3.00 \mathrm{~N}\) is applied tangential to the surface of the cylinder. You measure the angular displacement \(\theta-\theta_{0}\) of the cylinder as a function of the time \(t\) since the force was first applied. When you plot \(\theta-\theta_{0}\) (in radians) as a function of \(t^{2}\left(\right.\) in \(\left.\mathrm{s}^{2}\right),\) your data lie close to a straight line. If the slope of this line is \(16.0 \mathrm{rad} / \mathrm{s}^{2},\) what is the moment of inertia of the cylinder for rotation about the axle?

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix \(\mathrm{E}\) and the astronomical data in Appendix F.

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