/*! 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} Q7Q Make a sketch showing a situatio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Make a sketch showing a situation in which the torque due to a single force about some location is \(20\,\,{\rm{N}} \cdot {\rm{m}}\) in the positive \(z\) direction, whereas about another location the torque is \(10\,\,{\rm{N}} \cdot {\rm{m}}\) in the negative \(z\) direction.

Short Answer

Expert verified

The force that can cause an object to twist along an axis is measured as torque. In linear kinematics, force is what causes an object to accelerate.

Step by step solution

01

Definition of Torque

The force that can cause an object to twist along an axis is measured as torque. In linear kinematics, force is what causes an object to accelerate.

02

Concept about the Torque

The relation describes the torque operating on an object at a specific point.

\(\overrightarrow \tau = \overrightarrow r \times \overrightarrow F \)……. (1)

Here,

\(\overrightarrow r \to \)Position vector of the given location

\(\overrightarrow F \to \)Force vector acting at the location

03

Figure shows the position of force and vector to create a torque

The force and position vectors to create a torque at a given place are depicted in the figure:

04

Find the torque which acting about some location which is given in figure

Let unit vectors along X, Y, X axes are \(\widehat x,\widehat y,\widehat z\)respectively. In order to get torque acting about some location is, \(\tau = 20N \cdot m\)along the positive \(z\)direction, it is essential to consider, the position vector of that location is along the positive \(X\)axis\(\overrightarrow r = (2m)\widehat x\)and force vector acting along the positive \(Y\)axis, \(\overrightarrow F = (10N)\widehat y\)

Using equation (1), torque acting about some location is given by,

\(\begin{aligned}{}\overrightarrow \tau &= \overrightarrow r \times \overrightarrow F \\ &= (2m)\widehat x \times (10N)\widehat y\\ &= (2N \cdot m)(\widehat x \times \widehat y)\\ &= (2N \cdot m)\widehat z\,\,\,\,\,\,\,\,\,\,\,\,\,\,as\,(\widehat x \times \widehat y) = \widehat z)\end{aligned}\)

In order to get torque acting about another location is, \(\tau = 10N \cdot m\)along the negative \(z\)direction, it is essential to consider, the position vector of that location is along the negative \(X\)-axis \(\overrightarrow r = (1m)( - \widehat x)\)and same force vector acting along the positive \(Y\)axis, \(\overrightarrow F = (10N)\widehat y\).

Using equation (1), torque acting about another location is given by,

\(\begin{aligned}{}\overrightarrow \tau & = \overrightarrow r \times \overrightarrow F \\ &= (1m)( - \widehat x) \times (10N)\widehat y\\ &= (10N \cdot m)( - \widehat x \times \widehat y)\\ &= (10N \cdot m)( - \widehat z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,As( - \widehat x \times \widehat y) &= - \widehat z)\end{aligned}\)

Torque produced on positive and negative direction is\(2\,\,{\rm{N}}\,{\rm{m}}\) and \( - 10\,\,{\rm{N}}\,{\rm{m}}\) respectively.

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!

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

In figure two small objects each of mass m=0.3kgare connected by a lightweight rod of length d=1.5m.At a particular instant they have velocities whose magnitude are v1=38m/sand v2=60m/sand are subjected to external forces whose magnitudes are F1=41NandF2=26N. The distance role="math" localid="1668661918159" h=0.3m,and the distancew=0.7m.The system is moving in outer space. Assuming the usual coordinate system with+xto the right, +ytoward the top of the page, and +zout of the page toward you, calculated these quantities for this system:

(a) p→total,(b) v→CM, (c) L→tot,A, (d)L→rot,(e) L→transA, (f) P→totalat a time 0.23s after the initial time.

A rotating uniform-density disk of radius 0.6mis mounted in the vertical plane, as shown in Figure 11.88.The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 5kg.A lump of clay with mass0.4kgfalls and sticks to the outer edge of the wheel at the location ⟨-0.36,0.480,0⟩m,relative to an origin at the centre of the axle. Just before the impact the clay has speed 8m/s,and the disk is rotating clockwise with angular speed0.51radians/s.

(a) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the centerC?(As usual,xis to the right,yis up, andzis out of the screen, toward you.) (b) Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the centerC?(c) Just after the impact, what is the angular velocity (magnitude and direction) of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why? (A) There is no change because linear momentum is always conserved. (B) Some of the linear momentum is changed into angular momentum. (C) Some of the linear momentum is changed into energy. (D) The downward linear momentum decreases because the axle exerts an upward force.

Give an example of physical situation in which the angular momentum is zero yet the translational and rotational angular momenta are both non-zero.

A stationary bicycle wheel of radiusis mounted in the vertical plane on a horizontal low-friction axle (Figur The 11.43).Thewheel has mass,M all concentrated in the rim (the spokes have negligible mass). A lump of clay with mass m falls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has a speed v(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C (b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the centerCin terms of the angular speed of the wheel? (c) Just after the impact, what are the magnitude and direction of the angular velocity of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why?

A stationary bicycle wheel of radius 0.9mis mounted in the vertical plane (figure). The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. The wheel has mass all 4.8kgconcentrated in the rim (the spokes have negligible mass). A lump of clay with mass 0.5kgfalls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has speed 5m/sand the wheel is rotating clockwise with angular speed0.33rad/s.

(a) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? (As usual, is to the right, is up, and is out of the screen, towards you) (b) Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? (c) Just after the impact, what is the angular velocity (magnitude and direction) of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why? (1) The downward linear momentum decreases because the axle exerts into angular momentum. (3) some of the linear momentum is changed into energy. (4) There is no change because linear momentum is always conserved.

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.