/*! 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 8 A massless tether with a masses ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A massless tether with a masses tied to both ends rotates about a fixed axis through the center. Can the total acceleration of the tether/mass combination be zero if the angular velocity is constant?

Short Answer

Expert verified
No, the total acceleration of the tether/mass system cannot be zero, even if the angular velocity is constant. This is due to the presence of a non-zero centripetal acceleration which arises from the change in direction of the velocity vector of the tethered masses.

Step by step solution

01

Understand the significance of angular velocity

Angular velocity, represented as ω, is the rate of change of an angle with which a rigid body rotates. In this exercise, the angular velocity is constant. This signifies that the rate at which the tether/mass system is rotating isn't changing over time.
02

Consider the presence of centripetal acceleration

While the rotational speed may be constant, the direction of the velocity vector of the tethered masses changes with time. This change in direction leads to acceleration, more specifically centripetal acceleration directed towards the center of rotation.
03

Conclude whether total acceleration can be zero

Since there is a change in the direction of the velocity vector of the tethered masses (even if the magnitude of the rotational speed isn't changing), there is indeed a non-zero acceleration called centripetal acceleration. Hence, the total acceleration of the tether/mass combination cannot be zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Centripetal Acceleration
When a mass or object moves in a circle or a curved path, it experiences what is referred to as centripetal acceleration. This type of acceleration is always directed towards the center of the circle and is essential for keeping the object in circular motion. Imagine swinging a ball on a string in a horizontal circle overhead; the string pulls the ball toward the center, preventing it from flying off in a straight line, which is what it would naturally do due to inertia.

In physics, centripetal acceleration (\( a_c \) can be calculated using the formula \( a_c = \frac{v^2}{r} \) or \( a_c = \omega^2 \cdot r \) where \(v\) represents the linear velocity, \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path. Therefore, even with a constant angular velocity, the presence of a curved path necessitates a non-zero centripetal acceleration.
Rigid Body Rotation

Consider a figure skater performing a spin on ice. As the skater pulls her arms in, she spins faster; this is an example of rigid body rotation. A rigid body is a solid object that doesn’t deform or change shape as it moves—like the tether and masses in the exercise. In such cases, all parts of the body rotate around a fixed axis, and every point in the body maintains a constant distance from that axis.

During rigid body rotation, different points on the object have different linear speeds, but share the same angular velocity. This is because while the angle covered by all points in a given time is the same (due to the constant angular velocity), the distances from the axis are different, leading to varying paths and thus different linear speeds.
Constant Angular Velocity
If an object rotates at a constant angular velocity, it means its rate of rotation does not change over time. Angular velocity (\( \omega \) is measured in radians per second (rad/s) and provides a way to express the speed of rotation. For instance, a ferris wheel that takes a minute to complete a full revolution has a slower angular velocity compared to a spinning top that completes multiple revolutions in a second.

With a constant angular velocity, although the rate of rotation is steady, it is crucial to understand that it doesn't imply there's no acceleration. If there's a change in the direction of the velocity vector of the rotating object, as seen in the rotating tether with masses, centripetal acceleration will be present, and therefore, total acceleration cannot be zero despite the constant speed of rotation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A simple pendulum consists of a massless tether \(50 \mathrm{cm}\) in length connected to a pivot and a small mass of \(1.0 \mathrm{kg}\) attached at the other end. What is the torque about the pivot when the pendulum makes an angle of \(40^{\circ}\) with respect to the vertical?

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is the average angular acceleration in \(\operatorname{rad} / \mathrm{s}^{2} ?\) (b) What is the tangential acceleration of a point \(9.50 \mathrm{cm}\) from the axis of rotation? (c) What is the centripetal acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and multiples of \(g\) of this point at full rpm? (d) What is the total distance travelled by a point \(9.5 \mathrm{cm}\) from the axis of rotation of the ultracentrifuge?

Suppose when Earth was created, it was not rotating. However, after the application of a uniform torque after 6 days, it was rotating at 1 rev/day. (a) What was the angular acceleration during the 6 days? (b) What torque was applied to Earth during this period? (c) What force tangent to Earth at its equator would produce this torque?

Zorch, an archenemy of Rotation Man, decides to slow Earth's rotation to once per \(28.0 \mathrm{h}\) by exerting an opposing force at and parallel to the equator. Rotation Man is not immediately concerned, because he knows Zorch can only exert a force of \(4.00 \times 10^{7} \mathrm{N}\) (a little greater than a Saturn V rocket's thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Rotation Man time to devote to other villains.)

In the expression \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) can \(|\overrightarrow{\mathbf{r}}|\) ever be less than the lever arm? Can it be equal to the lever arm?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.