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20. Circular Motion

From Mechanics

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20. Circular Motion
 
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Key Points
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== '''Key Points''' ==
The diagram shows a particle
The diagram shows a particle
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The position vector for the particle is:
The position vector for the particle is:
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<math>\mathbf{r}=\left( r\cos \theta \right)\mathbf{i}+\left( r\sin \theta \right)\mathbf{j}</math>
<math>\mathbf{r}=\left( r\cos \theta \right)\mathbf{i}+\left( r\sin \theta \right)\mathbf{j}</math>
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The particle has angular speed
The particle has angular speed
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<math>t</math>:
<math>t</math>:
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<math>\mathbf{r}=\left( r\cos \omega t \right)\mathbf{i}+\left( r\sin \omega t \right)\mathbf{j}</math>
<math>\mathbf{r}=\left( r\cos \omega t \right)\mathbf{i}+\left( r\sin \omega t \right)\mathbf{j}</math>
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Differentiating the position vector gives the velocity:
Differentiating the position vector gives the velocity:
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<math>\mathbf{v}=\left( -r\omega \sin \omega t \right)\mathbf{i}+\left( r\omega \cos \omega t \right)\mathbf{j}</math>
<math>\mathbf{v}=\left( -r\omega \sin \omega t \right)\mathbf{i}+\left( r\omega \cos \omega t \right)\mathbf{j}</math>
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The magnitude of the velocity can now be found:
The magnitude of the velocity can now be found:
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<math>\begin{align}
<math>\begin{align}
& v=\sqrt{{{\left( -r\omega \sin \omega t \right)}^{2}}+{{\left( r\omega \cos \omega t \right)}^{2}}} \\
& 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}}({{\sin }^{2}}\omega t+{{\cos }^{2}}\omega t)} \\
 +
& \\
& =\sqrt{{{r}^{2}}{{\omega }^{2}}} \\
& =\sqrt{{{r}^{2}}{{\omega }^{2}}} \\
& =r\omega
& =r\omega
\end{align}</math>
\end{align}</math>
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The particle has speed,
The particle has speed,
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Differentiating the velocity gives the acceleration:
Differentiating the velocity gives the acceleration:
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<math>\begin{align}
<math>\begin{align}
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& =-{{\omega }^{2}}\mathbf{r}
& =-{{\omega }^{2}}\mathbf{r}
\end{align}</math>
\end{align}</math>
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This shows that the acceleration has magnitude
This shows that the acceleration has magnitude
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and
and
<math>r.</math>
<math>r.</math>
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<math>a=r{{\left( \frac{v}{r} \right)}^{2}}=\frac{{{v}^{2}}}{r}</math>
<math>a=r{{\left( \frac{v}{r} \right)}^{2}}=\frac{{{v}^{2}}}{r}</math>
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'''[[Example 20.1]]'''
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Example 20.1
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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.
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.
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Solution
+
'''Solution'''
The diagram shows the forces acting on the particle, its weight, the normal reaction from the surface and the friction.
The diagram shows the forces acting on the particle, its weight, the normal reaction from the surface and the friction.
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[[Image:E20.1.GIF]]
[[Image:E20.1.GIF]]
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- 
- 
- 
The acceleration of
The acceleration of
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The weight can be calculated first:
The weight can be calculated first:
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<math>W=0.03\times 9.8=0.294\text{ N}</math>
<math>W=0.03\times 9.8=0.294\text{ N}</math>
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As the vertical forces balance:
As the vertical forces balance:
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<math>R=\text{ }0.\text{294 N}</math>
<math>R=\text{ }0.\text{294 N}</math>
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The angular speed must be converted from rpm to
The angular speed must be converted from rpm to
<math>\text{rad }{{\text{s}}^{-1}}</math>.
<math>\text{rad }{{\text{s}}^{-1}}</math>.
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<math>\begin{align}
<math>\begin{align}
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Example 20.2
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'''[[Example 20.2]]'''
A car of mass 1200 kg travels round a bend with radius 20 metres at a constant speed of 12 <math>\text{m}{{\text{s}}^{-1}}</math>. 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.
A car of mass 1200 kg travels round a bend with radius 20 metres at a constant speed of 12 <math>\text{m}{{\text{s}}^{-1}}</math>. 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.
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Solution
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'''Solution'''
The diagram shows the forces acting on the car.
The diagram shows the forces acting on the car.
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\end{align}</math>
\end{align}</math>
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Resolving vertically:
Resolving vertically:
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<math>R=1200\times 9.8=11760\text{ N}</math>
<math>R=1200\times 9.8=11760\text{ N}</math>
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Then we can use the friction inequality
Then we can use the friction inequality
<math>F\le mR</math>, to give,
<math>F\le mR</math>, to give,
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<math>\begin{align}
<math>\begin{align}
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& \mu \ge \text{0}\text{.735 (to 3 sf)} \\
& \mu \ge \text{0}\text{.735 (to 3 sf)} \\
\end{align}</math>
\end{align}</math>
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So the least value of
So the least value of
<math>\mu </math>
<math>\mu </math>
is 0.735.
is 0.735.

Revision as of 11:25, 21 February 2010

       Theory          Exercises      


Key Points

The diagram shows a particle P

which moves with a circular path.

Image:T20.GIF

The position vector for the particle is:

r=rcosi+rsinj 

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

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

r=rcosti+rsintj 

Differentiating the position vector gives the velocity:

v=−rsinti+rcostj 

The magnitude of the velocity can now be found:

v=−rsint2+rcost2=r22(sin2t+cos2t)=r22=r

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

Differentiating the velocity gives the acceleration:

a=−r2costi+−r2sintj=−rcosti+rsintj=−2r

This shows that the acceleration has magnitude r2 and is directed towards the centre of the circle. We can write a=r2, but since v=r or =rv, the acceleration can be expressed in terms of v and r

a=rrv2=rv2 


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.

Image:E20.1.GIF

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

The weight can be calculated first:

W=00398=0294 N

As the vertical forces balance:

R= 0294 N

The angular speed must be converted from rpm to rad s−1.

=90 rpm=60902 rad s-1=3 rad s-1

Using F = ma in the radial direction with a=r2 gives:

F=00301(3)2=0267 N

Using F=R gives:

0267=0294=02940267=091 (to 2sf)


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.

Image:E20.2.GIF

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 9.8=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.

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