20. Cirkulär rörelse - Versionshistorik

Tek: flyttade 20. Circular Motion till 20. Cirkulär rörelse

2010-03-01T14:13:50Z

flyttade 20. Circular Motion till 20. Cirkulär rörelse

← Äldre version

Versionen från 1 mars 2010 kl. 14.13

Ian den 21 februari 2010 kl. 11.35

2010-02-21T11:35:19Z

← Äldre version		Versionen från 21 februari 2010 kl. 11.35
Rad 95:		Rad 95:
	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
Rad 117:		Rad 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>

Rad 126:		Rad 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>

Rad 154:		Rad 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 den 21 februari 2010 kl. 11.25

2010-02-21T11:25:24Z

← Äldre version		Versionen från 21 februari 2010 kl. 11.25
Rad 8:		Rad 8:


-	20. Circular Motion

-	Key Points	+	== '''Key Points''' ==

	The diagram shows a particle		The diagram shows a particle
Rad 21:		Rad 20:
	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
Rad 39:		Rad 36:
	<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,
Rad 65:		Rad 58:

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

	<math>\begin{align}		<math>\begin{align}
Rad 72:		Rad 64:
	& =-{{\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
Rad 84:		Rad 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
Rad 111:		Rad 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}
Rad 153:		Rad 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.
Rad 174:		Rad 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>
Rad 183:		Rad 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}
Rad 190:		Rad 165:
	& \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 den 22 november 2009 kl. 18.51

2009-11-22T18:51:51Z

← Äldre version		Versionen från 22 november 2009 kl. 18.51
Rad 9:		Rad 9:

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

Rad 15:		Rad 16:

	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:
Rad 97:		Rad 100:


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


Rad 157:		Rad 160:

	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 den 22 november 2009 kl. 18.09

2009-11-22T18:09:08Z

← Äldre version		Versionen från 22 november 2009 kl. 18.09
Rad 6:		Rad 6:
	\| 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]...

2009-11-16T18:34:38Z

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]...

Ny sida

__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]]}}
{{Not selected tab|[[20. Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
|}