20. Circular Motion - Revision history

Ian at 13:13, 27 April 2011

Ian — Wed, 27 Apr 2011 13:13:02 GMT

←Older revision		Revision as of 13:13, 27 April 2011
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-
-

	== '''Key Points''' ==		== '''Key Points''' ==

Ian at 15:12, 12 October 2010

Ian — Tue, 12 Oct 2010 15:12:43 GMT

←Older revision		Revision as of 15:12, 12 October 2010
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	The particle has angular speed		The particle has angular speed
-	<math>\omega </math>. This means that the angle	+	<math>\omega </math> which we assume is constant. This means that the angle
	<math>AOP</math>		<math>AOP</math>
	will increase by		will increase by
	<math>\omega </math>		<math>\omega </math>
	radians every second.		radians every second.
		+
		+	Variable angular speed lies outside the scope of this course.

	If the particle is at		If the particle is at
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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>
		+
		+	Using Newton’s second law, <math>F=ma</math>, gives the magnitude of the force needed to keep the particle in its circular path

Ian at 11:35, 21 February 2010

Ian — Sun, 21 Feb 2010 11:35:19 GMT

←Older revision		Revision as of 11:35, 21 February 2010
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	The weight can be calculated first:		The weight can be calculated first:

-	<math>W=0.03\times 9.8=0.294\text{ N}</math>	+	<math>W=0\textrm{.}03\times 9\textrm{.}8=0\textrm{.}294\text{ N}</math>

	As the vertical forces balance:		As the vertical forces balance:

-	<math>R=\text{ }0.\text{294 N}</math>	+	<math>R=\text{ }0\textrm{.}\text{294 N}</math>

	The angular speed must be converted from rpm to		The angular speed must be converted from rpm to
Line 117:		Line 117:

	<math>\begin{align}		<math>\begin{align}
-	& F=0.03\times 0.1\times {{(3\pi )}^{2}} \\	+	& F=0\textrm{.}03\times 0\textrm{.}1\times {{(3\pi )}^{2}} \\
-	& =0.267\text{ N}	+	& =0\textrm{.}267\text{ N}
	\end{align}</math>		\end{align}</math>

Line 126:		Line 126:

	<math>\begin{align}		<math>\begin{align}
-	& 0.267=0.294\mu \\	+	& 0\textrm{.}267=0\textrm{.}294\mu \\
-	& \mu =\frac{0.267}{0.294}=0.91\text{ (to 2sf)} \\	+	& \mu =\frac{0\textrm{.}267}{0\textrm{.}294}=0\textrm{.}91\text{ (to 2sf)} \\
	\end{align}</math>		\end{align}</math>

Line 154:		Line 154:
	Resolving vertically:		Resolving vertically:

-	<math>R=1200\times 9.8=11760\text{ N}</math>	+	<math>R=1200\times 9\textrm{.}8=11760\text{ N}</math>

Ian at 11:25, 21 February 2010

Ian — Sun, 21 Feb 2010 11:25:24 GMT

←Older revision		Revision as of 11:25, 21 February 2010
Line 8:		Line 8:


-	20. Circular Motion

-	Key Points	+	== '''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:

-
	<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>
-

	The particle has angular speed		The particle has angular speed
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	<math>t</math>:		<math>t</math>:

-
	<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>
-

	Differentiating the position vector gives the velocity:		Differentiating the position vector gives the velocity:

-
	<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>
-

	The magnitude of the velocity can now be found:		The magnitude of the velocity can now be found:

-
	<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>
-

	The particle has speed,		The particle has speed,
Line 65:		Line 58:

	Differentiating the velocity gives the acceleration:		Differentiating the velocity gives the acceleration:
-

	<math>\begin{align}		<math>\begin{align}
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	& =-{{\omega }^{2}}\mathbf{r}		& =-{{\omega }^{2}}\mathbf{r}
	\end{align}</math>		\end{align}</math>
-

	This shows that the acceleration has magnitude		This shows that the acceleration has magnitude
Line 84:		Line 75:
	and		and
	<math>r.</math>		<math>r.</math>
-
-

	<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>


-		+	'''[[Example 20.1]]'''
-	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.		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	+	'''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.
-

	[[Image:E20.1.GIF]]		[[Image:E20.1.GIF]]
-
-
-
-

	The acceleration of		The acceleration of
Line 111:		Line 94:

	The weight can be calculated first:		The weight can be calculated first:
-

	<math>W=0.03\times 9.8=0.294\text{ N}</math>		<math>W=0.03\times 9.8=0.294\text{ N}</math>
-

	As the vertical forces balance:		As the vertical forces balance:
-

	<math>R=\text{ }0.\text{294 N}</math>		<math>R=\text{ }0.\text{294 N}</math>
-

	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>.
-

	<math>\begin{align}		<math>\begin{align}
Line 153:		Line 131:


-	Example 20.2	+	'''[[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.

-	Solution	+	'''Solution'''

	The diagram shows the forces acting on the car.		The diagram shows the forces acting on the car.
Line 174:		Line 152:
	\end{align}</math>		\end{align}</math>

-
	Resolving vertically:		Resolving vertically:
-

	<math>R=1200\times 9.8=11760\text{ N}</math>		<math>R=1200\times 9.8=11760\text{ N}</math>
Line 183:		Line 159:
	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,
-

	<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>
-

	So the least value of		So the least value of
	<math>\mu </math>		<math>\mu </math>
	is 0.735.		is 0.735.

Ian at 18:51, 22 November 2009

Ian — Sun, 22 Nov 2009 18:51:51 GMT

←Older revision		Revision as of 18:51, 22 November 2009
Line 9:		Line 9:

	20. Circular Motion		20. Circular Motion
		+
	Key Points		Key Points

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	which moves with a circular path.		which moves with a circular path.
		+
		+	[[Image:T20.GIF]]

	The position vector for the particle is:		The position vector for the particle is:
Line 97:		Line 100:


-		+	[[Image:E20.1.GIF]]


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	The diagram shows the forces acting on the car.		The diagram shows the forces acting on the car.
		+
		+	[[Image:E20.2.GIF]]

	Using		Using

Ian at 18:09, 22 November 2009

Ian — Sun, 22 Nov 2009 18:09:08 GMT

←Older revision		Revision as of 18:09, 22 November 2009
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	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
		+
		+
		+	20. Circular Motion
		+	Key Points
		+
		+	The diagram shows a particle
		+	<math>P</math>
		+
		+	which moves with a circular path.
		+
		+	The position vector for the particle is:
		+
		+
		+	<math>\mathbf{r}=\left( r\cos \theta \right)\mathbf{i}+\left( r\sin \theta \right)\mathbf{j}</math>
		+
		+
		+	The particle has angular speed
		+	<math>\omega </math>. This means that the angle
		+	<math>AOP</math>
		+	will increase by
		+	<math>\omega </math>
		+	radians every second.
		+
		+	If the particle is at
		+	<math>A</math>
		+	when
		+	<math>t=0</math>, then
		+	<math>\theta =\omega t</math>. Using this, the position vector can be written in terms of
		+	<math>t</math>:
		+
		+
		+	<math>\mathbf{r}=\left( r\cos \omega t \right)\mathbf{i}+\left( r\sin \omega t \right)\mathbf{j}</math>
		+
		+
		+	Differentiating the position vector gives the velocity:
		+
		+
		+	<math>\mathbf{v}=\left( -r\omega \sin \omega t \right)\mathbf{i}+\left( r\omega \cos \omega t \right)\mathbf{j}</math>
		+
		+
		+	The magnitude of the velocity can now be found:
		+
		+
		+	<math>\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}</math>
		+
		+
		+	The particle has speed,
		+	<math>v</math>, which is given by
		+	<math>v=r\omega </math>. The velocity is directed along a tangent to the circle.
		+
		+	Differentiating the velocity gives the acceleration:
		+
		+
		+	<math>\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}</math>
		+
		+
		+	This shows that the acceleration has magnitude
		+	<math>r{{\omega }^{2}}</math>
		+	and is directed towards the centre of the circle. We can write
		+	<math>a=r{{\omega }^{2}}</math>, but since
		+	<math>v=r\omega </math>
		+	or
		+	<math>\omega =\frac{v}{r}</math>, the acceleration can be expressed in terms of
		+	<math>v</math>
		+	and
		+	<math>r.</math>
		+
		+
		+
		+	<math>a=r{{\left( \frac{v}{r} \right)}^{2}}=\frac{{{v}^{2}}}{r}</math>
		+
		+
		+
		+	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
		+	<math>P</math>
		+	is horizontal, so the resultant of the vertical forces must be zero.
		+
		+	The weight can be calculated first:
		+
		+
		+	<math>W=0.03\times 9.8=0.294\text{ N}</math>
		+
		+
		+	As the vertical forces balance:
		+
		+
		+	<math>R=\text{ }0.\text{294 N}</math>
		+
		+
		+	The angular speed must be converted from rpm to
		+	<math>\text{rad }{{\text{s}}^{-1}}</math>.
		+
		+
		+	<math>\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}</math>
		+
		+	Using
		+	<math>F\text{ }=\text{ }ma</math>
		+	in the radial direction with
		+	<math>a=r{{\omega }^{2}}</math>
		+	gives:
		+
		+	<math>\begin{align}
		+	& F=0.03\times 0.1\times {{(3\pi )}^{2}} \\
		+	& =0.267\text{ N}
		+	\end{align}</math>
		+
		+	Using
		+	<math>F=\mu R</math>
		+	gives:
		+
		+	<math>\begin{align}
		+	& 0.267=0.294\mu \\
		+	& \mu =\frac{0.267}{0.294}=0.91\text{ (to 2sf)} \\
		+	\end{align}</math>
		+
		+
		+	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.
		+
		+	Solution
		+
		+	The diagram shows the forces acting on the car.
		+
		+	Using
		+	<math>F=ma</math>
		+	radially, with
		+	<math>a=\frac{{{v}^{2}}}{r}</math>
		+	gives:
		+
		+	<math>\begin{align}
		+	& F=1200\times \frac{{{12}^{2}}}{20}\text{ } \\
		+	& =8640\text{ N}
		+	\end{align}</math>
		+
		+
		+	Resolving vertically:
		+
		+
		+	<math>R=1200\times 9.8=11760\text{ N}</math>
		+
		+
		+	Then we can use the friction inequality
		+	<math>F\le mR</math>, to give,
		+
		+
		+	<math>\begin{align}
		+	& 8640\le \mu \times 11760 \\
		+	& \mu \ge \frac{8640}{11760} \\
		+	& \mu \ge \text{0}\text{.735 (to 3 sf)} \\
		+	\end{align}</math>
		+
		+
		+	So the least value of
		+	<math>\mu </math>
		+	is 0.735.

Ian: New page: NOTOC {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | {{Selected tab|[[20. Circular Motion|Theory]...

Ian — Mon, 16 Nov 2009 18:34:38 GMT

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