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S5.3 Relations between angular and linear quantities

[A] Position

Key Point 5.7

If a body is rotated through an angle θ, a point a distance r from the rotation axis is moved in a circular arc of length

s=rθ

Graphic - No title - angtopos

Commentary

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Commentary

This applies only to rigid bodies.

The angle in radians, being defined as a ratio of two lengths is actually dimensionless ie we can say θ=1.5 radians or equally correctly, θ=1.5.

If θ=2π, s=2πr, the circumference of a circle of radius r

 

[B] Speed and period

Key Point 5.8

The speed, v of a point a distance r from the rotation axis of a body with angular velocity ω is

\[ v = r \frac{d \theta}{dt} = r \omega \]

The period of revolution is given by [distance]/[velocity]

\[ T = \frac{2 \pi r}{v} = \frac{2 \pi}{\omega} \]

Worked example

Worked example

Our sun is 2.3×104 light years from the centre of the Milky Way galaxy and is moving in a circle around that centre at a speed of 250 km/s.

  • [(a)] How long does it take the Sun to make one revolution about the galactic centre?
  • [(b)] How many revolutions has the Sun completed since it was formed 4.5×109 years ago?

Solution

  • [(a)] The period of revolution is given by T=2πr/v. The radius of orbit is given by the distance away in light years, multiplied by the distance light travels in one year:

    \begin{eqnarray*} %%%Was makeeqnarraystar; label was sun1 r &=& \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{2.3\times 10^4 \times c \times 365\times 24\times 3600} \\ &=& \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{2.17\times 10^{20}} m \end{eqnarray*}

    Hence

    \begin{eqnarray*} %%%Was makeeqnarraystar; label was sun2 T &=& { 2 \pi\times 2.17\times 10^{20}} / {2.5\times 10^5} \\ &=& 5.47\times 10^{15} s \\ &=& 1.7\times 10^8 years \end{eqnarray*}
  • [(b)] The number of revolutions, n is given by
    \begin{eqnarray*} %%%Was makeeqnarraystar; label was sun3 n &=& {4.5 \times 10^9} / {1.7 \times 10^8} \\ &=& 26 \end{eqnarray*}
 

[C] Velocity

Key Point 5.9

The velocity of a point $\vec{r}$ from the rotation axis is related to the angular velocity vector, $\vec{\omega}$ by the cross product (Key Point 0.4)

\[ \vec{v} = \vec{\omega} \times \vec{r} \]

Graphic - No title - omegav

[D] Acceleration

Key Point 5.10

The tangential component of the (linear) acceleration is

\[ a_t = r \frac{d \omega}{dt} = r \alpha. \]

The radial component of the (linear) acceleration, ar (Key Point 1.8) is

\[ a_r = \frac{v^2}{r} = r \omega^2 \]

Graphic - No title - angacc

Commentary

Commentary

The linear acceleration of a point on a rotating rigid body has in general, two components. The radially inwards component ar is present unless the angular velocity of the body is zero. The tangential component at is present unless the angular acceleration is zero. Thus KP 6.10 represents a generalisation KP 2.8 to the case of non zero angular acceleration.
 
TAngular Accelaration of a Fan
TAngular Acceleration on a spinning wheel
TAngular motion of a CD

Learning Resources

Textbook: HRW Chapter 10.5
Course Questions:
Self-Test Questions: