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Chapter 4: Problem 51

A rigid body is rotating with angular velocity \(6 \mathrm{rad} \mathrm{s}^{-1}\) about an axis in the direction of the vector \((3,-2,1)\) and passing through the point \(\mathrm{A}(3,-2,5)\). Find the linear velocity of the point \(\mathrm{P}(3,-2,1)\) on the body.

Short Answer

Expert verified
The linear velocity is \((-24, 48, 0)\, \mathrm{m}\, \mathrm{s}^{-1}\).

Step by step solution

01

Understand the Problem

We are given a rigid body rotating with an angular velocity of \(6 \ \mathrm{rad \, s}^{-1}\) about an axis defined by the vector \((3,-2,1)\). The axis passes through the point \(\mathrm{A}(3,-2,5)\). We need to find the linear velocity of the point \(\mathrm{P}(3,-2,1)\) on that body.
02

Determine the Vector from A to P

Find the vector from point \(\mathrm{A}\) to point \(\mathrm{P}\) by subtracting the coordinates of \(\mathrm{A}\) from those of \(\mathrm{P}\). This yields: \[\mathbf{AP} = (3-3, -2-(-2), 1-5) = (0, 0, -4).\]
03

Calculate the Linear Velocity

The linear velocity \(\mathbf{v}\) of a point on a rotating body is given by the cross product of the angular velocity vector \(\mathbf{\omega}\) and the position vector \(\mathbf{AP}\). First, express \(\mathbf{\omega}\) as a vector using the direction vector \((3, -2, 1)\): \[\mathbf{\omega} = 6 \left( \frac{3}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right).\] Now calculate the cross product \(\mathbf{v} = \mathbf{\omega} \times \mathbf{AP}\):\[\mathbf{v} = \left( 6 \times \frac{3}{\sqrt{14}}, 6 \times \frac{-2}{\sqrt{14}}, 6 \times \frac{1}{\sqrt{14}} \right) \times (0, 0, -4) = \left( -\frac{24}{\sqrt{14}}, \frac{48}{\sqrt{14}}, 0 \right).\]
04

Simplify the Result

To simplify the expression for the linear velocity, multiply through by \(\sqrt{14}\) to clear the fraction:\[\mathbf{v} = \left( -24, 48, 0 \right)\, \mathrm{m}\, \mathrm{s}^{-1}.\]
05

State the Final Answer

The linear velocity vector of point \(\mathrm{P}(3, -2, 1)\) is \((-24, 48, 0)\, \mathrm{m}\, \mathrm{s}^{-1}\).

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Key Concepts

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

Rigid Body Rotation
Imagine holding a solid object in your hand and spinning it around. This action is known as rigid body rotation. It refers to the movement of an object where all parts rotate around a specific axis without distorting. Rigid bodies are often considered in physics because they do not deform, making them easier to analyze in rotational dynamics.
  • The axis of rotation can be inside or outside the body.
  • Every point in the object follows a circular path around the axis.
  • Rigid body rotation is a key concept when analyzing rotating objects like wheels, discs, and planets.
In analyzing these rotations mathematically, we often use vectors to describe the axis and the rotational qualities of the body.
Angular Velocity
The speed at which a rigid body rotates is described by its angular velocity. It measures how quickly the angle changes as the body rotates. Angular velocity is a vector quantity, which means it has both a magnitude and a direction. The magnitude tells how fast the rotation occurs, while the direction follows the right-hand rule around the axis of rotation.
  • It is commonly measured in radians per second (\( ext{rad/s}\)).
  • The angular velocity vector points along the axis of rotation.
  • It is crucial in calculations involving centripetal force and torque.
In our problem, the rigid body has an angular velocity of \(6 \, \mathrm{rad/s}\) along the vector \((3, -2, 1)\), indicating a high-speed rotation around this direction.
Cross Product
In the realm of vectors, the cross product is a special operation that results in a new vector. It is used to find the perpendicular vector to two given vectors. In the context of rotational movement, the cross product helps determine the linear velocity of a point on the rotating body.
  • The cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\mathbf{a} \times \mathbf{b}\).
  • This operation is crucial in physics when calculating moments and forces.
  • In our solution, the cross product of \(\mathbf{\omega}\) (angular velocity vector) and \(\mathbf{AP}\) (position vector from the axis to the point) gives the linear velocity \(\mathbf{v}\).
The result is a vector that shows the speed and direction of point \(P\) as the body rotates.
Position Vector
A position vector describes the location of a point in space relative to a reference. In this exercise, we use the position vector to express the location between a given point on a rotating body and the axis of rotation.
  • A position vector is an arrow from one point to another, denoted by coordinate differences.
  • For point \(P(3, -2, 1)\) relative to axis point \(A(3, -2, 5)\), the position vector is \((0, 0, -4)\).
  • This vector helps us calculate how fast and in what direction a point moves when a body rotates.
Understanding position vectors allows us to determine relationships between points and axes in rotational systems.

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Most popular questions from this chapter

A cyclist travelling east at 8 kilometres per hour finds that the wind appears to blow directly from the north. On doubling her speed it appears to blow from the north-east. Find the actual velocity of the wind.

Find the shortest distance between the two lines $$ \boldsymbol{r}=(4,-2,3)+t(2,1,-1) $$ and $$ \boldsymbol{r}=(-7,-2,1)+s(3,2,1) $$

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