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Chapter 10: Problem 28

(II) A wheel of diameter 27.0 \(\mathrm{cm}\) is constrained to rotate in the \(x y\) plane, about the \(z\) axis, which passes through its center. A force \(\vec{\mathbf{F}}=(-31.0 \hat{\mathbf{i}}+43.4 \hat{\mathbf{j}}) \mathrm{N}\) acts at a point on the edge of the wheel that lies exactly on the \(x\) axis at a particular instant. What is the torque about the rotation axis at this instant?

Short Answer

Expert verified
The torque is 5.859 Nm about the z-axis.

Step by step solution

01

Understand the Problem

The problem involves calculating the torque exerted on a wheel by a force. The wheel rotates about the z-axis, and we are given the force vector acting at a specific point on the edge of the wheel. Torque is the rotational equivalent of force and is given by the cross product of the lever arm and the force.
02

Determine the Lever Arm

Since the wheel's diameter is 27.0 cm, the radius (and thus the lever arm, \( \vec{r} \)) is half of the diameter, which is 13.5 cm or 0.135 m. At the instant described, this point on the edge of the wheel lies on the x-axis, meaning \( \vec{r} = 0.135 \, \hat{\mathbf{i}} \).
03

Find the Position and Force Vectors

The position vector \( \vec{r} \) from the center to the point of force application is \( 0.135 \, \hat{\mathbf{i}} \) m. The force vector is given as \( \vec{\mathbf{F}} = (-31.0 \, \hat{\mathbf{i}} + 43.4 \, \hat{\mathbf{j}}) \, \text{N} \).
04

Calculate the Torque using Cross Product

Torque \( \vec{\tau} \) is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). To compute this, set up the determinant including unit vectors and the vectors: \[ \vec{\tau} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 0.135 & 0 & 0 \ -31.0 & 43.4 & 0 \end{vmatrix} \] Calculating the determinant:\[ \vec{\tau} = \hat{\mathbf{k}}(0.135 \times 43.4 - 0 \times (-31.0)) \]This simplifies to:\[ \vec{\tau} = 0.135 \times 43.4 \, \hat{\mathbf{k}} \]
05

Complete the Calculation

Perform the multiplication: \[ 0.135 \times 43.4 = 5.859 \] Therefore, the torque \( \vec{\tau} = 5.859 \, \hat{\mathbf{k}} \, \text{Nm} \). The direction of this torque is along the \( \hat{\mathbf{k}} \) axis, which is consistent with it acting about the z-axis.

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Key Concepts

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

Cross Product
The cross product is a mathematical operation used to determine the torque in this context. The cross product of two vectors results in a third vector which is perpendicular to the plane formed by the first two. In easier terms, it helps us find the direction and magnitude of the twisting force, or torque, that is applied to an object.

When talking about torque, we use the cross product of the lever arm vector and the force vector. The formula is:
  • The torque \( \vec{\tau} \) is determined by \( \vec{\tau} = \vec{r} \times \vec{F} \).
  • The symbol \( \times \) represents the cross product.
This calculation helps us understand how the force is affecting the wheel's rotation. By solving the cross product, we find that the direction of the torque is along the z-axis, emphasizing that the wheel's rotation happens around the z-axis as well.
Lever Arm
The lever arm is a crucial element in calculating torque. It is essentially the radius or distance from the axis of rotation to where the force is applied. In simpler words, imagine the arm of a wrench; the lever arm is the length of the wrench that you apply force to turn a bolt.

In our problem:
  • The diameter of the wheel is given as 27.0 cm, so the lever arm, which is the radius, is half of this length.
  • This computes to 13.5 cm or converted to meters, 0.135 m.
  • The specific point on the wheel where force is applied lies on the x-axis, meaning the lever arm vector is \( 0.135 \, \hat{\mathbf{i}} \).
This straightforward concept helps establish the position from which rotational effects like torque begin.
Force Vector
The force vector is a representation of both the magnitude and direction of the force being applied. It tells us not just how strong the force is, but also in which direction it is pushing or pulling.

For the given wheel:
  • The force vector is given as \( \vec{\mathbf{F}} = (-31.0 \, \hat{\mathbf{i}} + 43.4 \, \hat{\mathbf{j}}) \, \text{N} \).
  • This indicates two components: one of \(-31.0\) N along the negative x-axis and another of \(43.4\) N along the positive y-axis.
By understanding both components of this vector, we can better visualize how the force impacts the wheel, not causing it to move straight but rather to rotate.
Rotation Axis
The rotation axis is an imaginary line that acts as a pivot around which an object rotates. In the context of torque and rotation, it is crucial to identify this axis as it determines how and where the forces will cause the object to turn.

For the wheel in this problem:
  • The rotation takes place around the z-axis.
  • This axis passes through the center of the wheel and is perpendicular to the plane in which the wheel lies.
  • With the force and torque calculated, we ascertain that the wheel will primarily rotate about this z-axis.
Knowing the rotation axis allows us to predict the direction and nature of the wheel's movement when a force acts upon it.

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Most popular questions from this chapter

(III) An Atwood's machine consists of two masses, \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) which are connected by a massless inelastic cord that passes over a pulley, Fig. \(57 .\) If the pulley has radius \(R\) and moment of inertia \(I\) about its axle, determine the acceleration of the masses \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}},\) and compare to the situation in which the moment of inertia of the pulley is ignored. [Hint: The tensions \(F_{\mathrm{TA}}\) and \(F_{\mathrm{TB}}\) are not equal. With the Atwood machine, assume \(I=0\) for the pulley.]

(a) A grinding wheel \(0.35 \mathrm{~m}\) in diameter rotates at 2500 rpm. Calculate its angular velocity in rad/s. (b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?

A wheel of diameter \(27.0 \mathrm{~cm}\) is constrained to rotate in the \(x y\) plane, about the \(z\) axis, which passes through its center. A force \(\overrightarrow{\mathbf{F}}=(-31.0 \hat{\mathbf{i}}+43.4 \hat{\mathbf{j}}) \mathbf{N}\) acts at a point on the edge of the wheel that lies exactly on the \(x\) axis at a particular instant. What is the torque about the rotation axis at this instant?

(I) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (a) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. [Assume the Earth is a uniform sphere with \(\operatorname{mass}=6.0 \times 10^{24} \mathrm{kg}\) , radius \(=6.4 \times 10^{6} \mathrm{m},\) and is \(1.5 \times 10^{8} \mathrm{km}\) from the Sun.

(II) A solid rubber ball rests on the floor of a railroad car when the car begins moving with acceleration \(a\) . Assuming the ball rolls without slipping, what is its acceleration relative to \((a)\) the car and \((b)\) the ground?
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