20. Cirkulär rörelse

Förberedande Mekanik

Nuvarande version

Key Points

The diagram shows a particle \displaystyle P

which moves with a circular path.

The position vector for the particle is:

\displaystyle \mathbf{r}=\left( r\cos \theta \right)\mathbf{i}+\left( r\sin \theta \right)\mathbf{j}

The particle has angular speed \displaystyle \omega . This means that the angle \displaystyle AOP will increase by \displaystyle \omega radians every second.

If the particle is at \displaystyle A when \displaystyle t=0, then \displaystyle \theta =\omega t. Using this, the position vector can be written in terms of \displaystyle t:

\displaystyle \mathbf{r}=\left( r\cos \omega t \right)\mathbf{i}+\left( r\sin \omega t \right)\mathbf{j}

Differentiating the position vector gives the velocity:

\displaystyle \mathbf{v}=\left( -r\omega \sin \omega t \right)\mathbf{i}+\left( r\omega \cos \omega t \right)\mathbf{j}

The magnitude of the velocity can now be found:

\displaystyle \begin{align} & v=\sqrt{{{\left( -r\omega \sin \omega t \right)}^{2}}+{{\left( r\omega \cos \omega t \right)}^{2}}} \\ & \\ & =\sqrt{{{r}^{2}}{{\omega }^{2}}({{\sin }^{2}}\omega t+{{\cos }^{2}}\omega t)} \\ & \\ & =\sqrt{{{r}^{2}}{{\omega }^{2}}} \\ & =r\omega \end{align}

The particle has speed, \displaystyle v, which is given by \displaystyle v=r\omega . The velocity is directed along a tangent to the circle.

Differentiating the velocity gives the acceleration:

\displaystyle \begin{align} & \mathbf{a}=\left( -r{{\omega }^{2}}\cos \omega t \right)\mathbf{i}+\left( -r{{\omega }^{2}}\sin \omega t \right)\mathbf{j} \\ & =-\omega \left( \left( r\cos \omega t \right)\mathbf{i}+\left( r\sin \omega t \right)\mathbf{j} \right) \\ & =-{{\omega }^{2}}\mathbf{r} \end{align}

This shows that the acceleration has magnitude \displaystyle r{{\omega }^{2}} and is directed towards the centre of the circle. We can write \displaystyle a=r{{\omega }^{2}}, but since \displaystyle v=r\omega or \displaystyle \omega =\frac{v}{r}, the acceleration can be expressed in terms of \displaystyle v and \displaystyle r.

\displaystyle a=r{{\left( \frac{v}{r} \right)}^{2}}=\frac{{{v}^{2}}}{r}

Example 20.1

A coin of mass 30 grams is placed on a turntable which rotates at 90 rpm. The coin is at a distance of 10 cm from the centre of the turntable. Find the magnitude of the coefficient of friction between the coin and the turntable, if the coin is on the point of slipping.

Solution

The diagram shows the forces acting on the particle, its weight, the normal reaction from the surface and the friction.

The acceleration of \displaystyle P is horizontal, so the resultant of the vertical forces must be zero.

The weight can be calculated first:

\displaystyle W=0\textrm{.}03\times 9\textrm{.}8=0\textrm{.}294\text{ N}

As the vertical forces balance:

\displaystyle R=\text{ }0\textrm{.}\text{294 N}

The angular speed must be converted from rpm to \displaystyle \text{rad }{{\text{s}}^{-1}}.

\displaystyle \begin{align} & \omega =90\text{ rpm} \\ & =\frac{90\times 2\pi }{60}\text{ rad }{{\text{s}}^{\text{-1}}} \\ & =\text{3}\pi \text{ rad }{{\text{s}}^{\text{-1}}} \end{align}

Using \displaystyle F\text{ }=\text{ }ma in the radial direction with \displaystyle a=r{{\omega }^{2}} gives:

\displaystyle \begin{align} & F=0\textrm{.}03\times 0\textrm{.}1\times {{(3\pi )}^{2}} \\ & =0\textrm{.}267\text{ N} \end{align}

Using \displaystyle F=\mu R gives:

\displaystyle \begin{align} & 0\textrm{.}267=0\textrm{.}294\mu \\ & \mu =\frac{0\textrm{.}267}{0\textrm{.}294}=0\textrm{.}91\text{ (to 2sf)} \\ \end{align}

Example 20.2

A car of mass 1200 kg travels round a bend with radius 20 metres at a constant speed of 12 \displaystyle \text{m}{{\text{s}}^{-1}}. Find the magnitude of the friction force acting on the car. Find the minimum possible value of the coefficient of friction between the tyres and the road.

Solution

The diagram shows the forces acting on the car.

Using \displaystyle F=ma radially, with \displaystyle a=\frac{{{v}^{2}}}{r} gives:

\displaystyle \begin{align} & F=1200\times \frac{{{12}^{2}}}{20}\text{ } \\ & =8640\text{ N} \end{align}

Resolving vertically:

\displaystyle R=1200\times 9\textrm{.}8=11760\text{ N}

Then we can use the friction inequality \displaystyle F\le mR, to give,

\displaystyle \begin{align} & 8640\le \mu \times 11760 \\ & \mu \ge \frac{8640}{11760} \\ & \mu \ge \text{0}\text{.735 (to 3 sf)} \\ \end{align}

So the least value of \displaystyle \mu is 0.735.