Circular Motion, Centripetal Force & Gravitation
Position of a Point on a Circle
A point \(P\) moves on a circle of radius \(r\).
Coordinates:
\[
x = r\cos\theta, \qquad y = r\sin\theta
\]
Position vector:
\[
\vec{r} = r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}
\]
Radial and Transverse Directions
The point is set in motion in an anticlockwise direction.
Its position at time t seconds is
\[
x(t) = r\cos\theta(t) \text{
, }
y(t) = r\sin\theta(t)
\]
\[
\mathbf{r}(t)
= x(t)\,\mathbf{i} + y(t)\,\mathbf{j}
\]
\[
\mathbf{r}(t)
= r\cos\theta(t)\,\mathbf{i}
+ r\sin\theta(t)\,\mathbf{j}
\]
where i and j are unit vectors in the positive x and y axes and θ is the anticlockwise angle between the point and the x axis.
The length of the radius is not dependant upon the time, since the point is moving around a circle.
PT gives the direction of travel, or traverse direction of the particle. This is tangential to the circle .
PN is the direction radially outwards, or radial,from the centre of the circle.
The angle between PN and PT is 90°.
Since the particle is moving, the velocity vector of the radial is found :
Velocity
\[
\dot{\mathbf{r}}(t)
= \dot{x}(t)\,\mathbf{i}
+ \dot{y}(t)\,\mathbf{j}
\]
so
\[
\dot{x}(t)
= -r\,\dot{\theta}\,\sin\theta(t)
\]
\[
\dot{y}(t)
= r\,\dot{\theta}\,\cos\theta(t)
\]
(Don't forget the chain rule!)
\[
\dot{\mathbf{r}}(t)
= \dot{x}(t)\,\mathbf{i}
+ \dot{y}(t)\,\mathbf{j}
\]
\[
\dot{\mathbf{r}}(t)
= -r\,\dot{\theta}\,\sin\theta(t)\,\mathbf{i}
+ r\,\dot{\theta}\,\cos\theta(t)\,\mathbf{j}
\]
Radial component:
\[
\dot{x}(t)\cos\theta(t)
+ \dot{y}(t)\sin\theta(t)
\]
\[
= -r\dot{\theta}\,\sin\theta(t)\cos\theta(t)
+ r\dot{\theta}\,\cos\theta(t)\sin\theta(t)
\]
\[
= 0
\]
There is no radial component of velocity.
Transverse component:
\[
-\dot{x}(t)\,\sin\theta(t)
+ \dot{y}(t)\,\cos\theta(t)
\]
\[
= -\bigl(-r\dot{\theta}\,\sin\theta(t)\bigr)\sin\theta(t)
+ r\dot{\theta}\,\cos\theta(t)\cos\theta(t)
\]
\[
= r\dot{\theta}\,\sin^{2}\theta(t)
+ r\dot{\theta}\,\cos^{2}\theta(t)
\]
\[
= r\dot{\theta}\,\bigl(\sin^{2}\theta(t)
+ \cos^{2}\theta(t)\bigr)
\]
\[
= r\dot{\theta}
\]
Acceleration
\[
\ddot{\mathbf{r}}(t)
= \ddot{x}(t)\,\mathbf{i}
+ \ddot{y}(t)\,\mathbf{j}
\]
\[
= \frac{d}{dt}
\Bigl(
-r\dot{\theta}\,\sin\theta(t)\,\mathbf{i}
+ r\dot{\theta}\,\cos\theta(t)\,\mathbf{j}
\Bigr)
\]
\[
=
\Bigl(
-r\ddot{\theta}\,\sin\theta(t)
- r\dot{\theta}^{2}\cos\theta(t)
\Bigr)\mathbf{i}
+
\Bigl(
r\ddot{\theta}\,\cos\theta(t)
- r\dot{\theta}^{2}\,\sin\theta(t)
\Bigr)\mathbf{j}
\]
\[
=
-r\Bigl(
\ddot{\theta}\,\sin\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)
\Bigr)\mathbf{i}
+
r\Bigl(
\ddot{\theta}\,\cos\theta(t)
- \dot{\theta}^{2}\,\sin\theta(t)
\Bigr)\mathbf{j}
\]
\[
=
-r\dot{\theta}^{2}
\bigl(\cos\theta(t)\,\mathbf{i}
+ \sin\theta(t)\,\mathbf{j}\bigr)
+
r\ddot{\theta}
\bigl(-\sin\theta(t)\,\mathbf{i}
+ \cos\theta(t)\,\mathbf{j}\bigr)
\]
\[
\ddot{x}(t)\,\mathbf{i}
= -r\Bigl(
\ddot{\theta}\,\sin\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)
\Bigr)\mathbf{i}
\]
\[
\ddot{x}(t)
= -r\Bigl(
\ddot{\theta}\,\sin\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)
\Bigr)
\]
\[
\ddot{y}(t)\,\mathbf{j}
= r\Bigl(
\ddot{\theta}\,\cos\theta(t)
- \dot{\theta}^{2}\,\sin\theta(t)
\Bigr)\mathbf{j}
\]
\[
\ddot{y}(t)
= r\Bigl(
\ddot{\theta}\,\cos\theta(t)
- \dot{\theta}^{2}\,\sin\theta(t)
\Bigr)
\]
Radial acceleration:
\[
\ddot{\mathbf{r}}(t)
= \ddot{x}(t)\cos\theta(t)\,\mathbf{i}
+ \ddot{y}(t)\sin\theta(t)\,\mathbf{j}
\]
\[
= -r\Bigl(
\ddot{\theta}\,\sin\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)
\Bigr)\cos\theta(t)
\]
\[
\quad
+\, r\Bigl(
\ddot{\theta}\,\cos\theta(t)
- \dot{\theta}^{2}\,\sin\theta(t)
\Bigr)\sin\theta(t)
\]
\[
= -r\ddot{\theta}\,\sin\theta(t)\cos\theta(t)
- r\dot{\theta}^{2}\cos^{2}\theta(t)
+ r\ddot{\theta}\,\cos\theta(t)\sin\theta(t)
- r\dot{\theta}^{2}\sin^{2}\theta(t)
\]
\[
= -r\dot{\theta}^{2}
\Bigl(
\cos^{2}\theta(t)
+ \sin^{2}\theta(t)
\Bigr)
\]
\[
= -r\dot{\theta}^{2}
\]
The negative value shows that radial acceleration is towards the centre.
Transverse acceleration:
\[
-\ddot{x}(t)\,\sin\theta(t)
+ \ddot{y}(t)\,\cos\theta(t)
\]
\[
= -\Bigl(
-r\bigl(\ddot{\theta}\sin\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)\bigr)
\Bigr)\sin\theta(t)
\]
\[
\quad
+\, r\Bigl(
\ddot{\theta}\cos\theta(t)
- \dot{\theta}^{2}\sin\theta(t)
\Bigr)\cos\theta(t)
\]
\[
= r\Bigl(
\ddot{\theta}\sin^{2}\theta(t)
+ \dot{\theta}^{2}\cos\theta(t)\sin\theta(t)
\Bigr)
+ r\Bigl(
\ddot{\theta}\cos^{2}\theta(t)
- \dot{\theta}^{2}\sin\theta(t)\cos\theta(t)
\Bigr)
\]
\[
= r\ddot{\theta}\bigl(\sin^{2}\theta(t)
+ \cos^{2}\theta(t)\bigr)
\]
\[
= r\ddot{\theta}
\]
\[
\text{Angular velocity}
\]
\[
\dot{\theta} = \omega
\]
\[
\text{Angular speed}
\]
\[
|\omega|
\]
The transverse velocity component \(
v = r\dot{\theta} = r\omega
\) can be referred to as the velocity, since the radial component is zero.
This leads to \( v = r\omega = \dot{x} \)
and
\[
a_{\text{transverse}}
= r\ddot{\theta}
\]
\[
= r\dot{\omega}
\]
Centripetal Acceleration
This is acceleration towards the the center of the circle.
\[
a_{\text{radial}}
= -r\,\dot{\theta}^{2}
\]
\[
= -r\,\omega^{2}
\]
\[
= -r\left(\frac{v}{r}\right)^{2}
\]
\[
= -\frac{v^{2}}{r}
\]
So the equation for centripetal acceleration is
Period of revolution: the time taken for one complete revolution.
\[
T = \frac{2\pi r}{v}
\]
Summary
Angular motion
\[
\theta(t) = \omega t + \varepsilon
\]
\[
\omega = \frac{d\theta}{dt}
\]
\[
T = \frac{2\pi}{\omega},
\quad
f = \frac{1}{T},
\quad
\omega = 2\pi f
\]
Velocity
\[
\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}
\]
\[
\mathbf{v}(t)
= -r\omega\sin\theta(t)\,\mathbf{i}
+ r\omega\cos\theta(t)\,\mathbf{j}
\]
\[
v = \|\mathbf{v}\| = r\omega
\]
Acceleration (centripetal)
\[
\mathbf{a}(t) = \frac{d\mathbf{v}}{dt}
\]
\[
\mathbf{a}(t)
= -r\omega^{2}\cos\theta(t)\,\mathbf{i}
- r\omega^{2}\sin\theta(t)\,\mathbf{j}
= -\omega^{2}\mathbf{r}(t)
\]
\[
a = \|\mathbf{a}\|
= \frac{v^{2}}{r}
= r\omega^{2}
\]
Projection as SHM
\[
x(t) = r\cos(\omega t + \varepsilon)
\]
\[
\ddot{x} = -\omega^{2}x
\]
Uniform Circular Motion
Newton's Laws
A particle moving in a circular path must be subject to forces - due to Newton's First Law.
There is no transverse acceleration, but Newton's Second Law shows that there is a force directed towards the circle's centre:
\[
F = ma
\]
\[
= m r \omega^{2}
\]
\[
= \frac{mv^{2}}{r}
\]
Banking an Aircraft
An aircraft is banking .
Applying a force diagram:
Here, the normal reaction R represents the lift acting on the aircraft.
Examining the horizontal component: \( F = R\sin\theta \)
Applying Newton's 2nd Law:
\[
F = ma
\]
\[R\sin \theta
= \frac{Mv^{2}}{r}
\]
Where v is the aircraft's true airspeed , M its mass and θ its angle of bank. The radius of turn is r.
\[R
= \frac{Mv^{2}}{r\sin \theta}
\]
Now looking at the vertical component: \( R \cos \theta = Mg \)
so
\[R
= \frac{Mg}{\cos \theta}
\]
Equating these
\[
R
= \frac{Mg}{\cos\theta}
= \frac{Mv^{2}}{r\sin\theta}
\]
\[
\frac{Mg}{\cos\theta}
= \frac{Mv^{2}}{r\sin\theta}
\]
\[
v^{2}
= \frac{gr\sin\theta}{\cos\theta}
\]
\[
v^{2}
= gr\tan\theta
\]
\[
v = \sqrt{gr\tan\theta}
\]
also
\[
\frac{Mg}{\cos\theta}
= \frac{Mv^{2}}{r}
\]
\[
r
= \frac{v^{2}\cos\theta}{g\sin\theta}
\]
\[
r
= \frac{v^{2}\cot\theta}{g}
\]
\[
r
= \frac{v^{2}}{g\tan\theta}
\]
The radius of the turn is proportional to the square of the true airspeed.
Horizontal component of lift provides centripetal force:
\[
R\sin\theta = \frac{mv^2}{r}
\]
Vertical component balances weight:
\[
R\cos\theta = mg
\]
Dividing:
\[
\tan\theta = \frac{v^2}{rg}
\]
Banked Motion (Cars)
A car is driving at a constant velocity around a banked race track.
The car experiences centripetal acceleration, so has a horizontal centripetal force:
Drawing a force diagram gives
If there is no tendency to slip, then there is no frictional force between the tyres and the road.
Resolving horizontally
\[
R\sin\theta = \frac{mv^2}{r}
\]
Resolving vertically,
\[
R\cos\theta = mg
\]
Equating gives the angle of inclination of the bank :
\[
R
= \frac{Mg}{\cos\theta}
= \frac{Mv^{2}}{r\sin\theta}
\]
\[
\frac{Mg}{\cos\theta}
= \frac{Mv^{2}}{r\sin\theta}
\]
\[
\frac{\sin\theta}{\cos\theta}
= \frac{Mv^{2}}{rMg}
\]
\[
\tan\theta
= \frac{v^{2}}{rg}
\]
\[
\theta
= \tan^{-1}\!\left(\frac{v^{2}}{rg}\right)
\]
If there is a tendency to slip, then there is a frictional force between the tyres and the road.
Resolving horizontally
\[
\mu R\cos\theta
+ R\sin\theta
= \left(\frac{Mv}{r}\right)^{2}
\]
Resolving vertically,
\[
R\cos\theta
= Mg + \mu R\sin\theta
\]
Rearranging these
\[
\mu R\cos\theta
+ R\sin\theta
= \left(\frac{Mv}{r}\right)^{2}
\]
\[
\frac{rR\big(\mu\cos\theta + \sin\theta\big)}{v^{2}}
= M
\]
and
\[
R\cos\theta
= Mg + \mu R\sin\theta
\]
\[
\frac{R\big(\cos\theta - \mu\sin\theta\big)}{g}
= M
\]
so
\[
\frac{R\big(\cos\theta - \mu\sin\theta\big)}{g}
=
\frac{rR\big(\mu\cos\theta + \sin\theta\big)}{v^{2}}
\]
\[
\frac{\cos\theta - \mu\sin\theta}{g}
=
\frac{r\big(\mu\cos\theta + \sin\theta\big)}{v^{2}}
\]
\[
\frac{\cos\theta - \mu\sin\theta}{\mu\cos\theta + \sin\theta}
=
\frac{gr}{v^{2}}
\]
so
\[
\frac{\cos\theta - \mu\sin\theta}{\mu\cos\theta + \sin\theta}
= \frac{gr}{v^{2}}
\]
\[
\frac{\cos\theta}{\cos\theta}
\times
\frac{
\left(\dfrac{\cos\theta - \mu\sin\theta}{\cos\theta}\right)
}{
\left(\dfrac{\mu\cos\theta + \sin\theta}{\cos\theta}\right)
}
= \frac{gr}{v^{2}}
\]
\[
\frac{1 - \mu\tan\theta}{\mu + \tan\theta}
= \frac{gr}{v^{2}}
\]
giving
\[
\frac{1 - \mu\tan\theta}{\mu + \mu\tan\theta}
= \frac{gr}{v^{2}}
\]
\[
v^{2}
= \frac{gr\big(\mu + \mu\tan\theta\big)}
{1 - \mu\tan\theta}
\]
\[
v
= \sqrt{
\frac{gr\big(\mu + \mu\tan\theta\big)}
{1 - \mu\tan\theta}
}
\]
and
\[
\frac{1 - \mu\tan\theta}{\mu + \mu\tan\theta}
= \frac{gr}{v^{2}}
\]
\[
r
= \frac{v^{2}\big(1 - \mu\tan\theta\big)}
{g\big(\mu + \mu\tan\theta\big)}
\]
If no friction:
\[
\tan\theta = \frac{v^2}{rg}
\]
With friction:
\[
\frac{v^2}{r} = g\frac{\sin\theta \pm \mu\cos\theta}{\cos\theta \mp \mu\sin\theta}
\]
Conical Pendulum
In the diagram below, a bob of mass m Kg is suspended from point A by a light inextensible string of length L. The bob is set to rotate with a constant speed in a circular motion of radius r about a point which is h m vertically below A, so that the string traces out a cone.
The tension in the string, T, splits into horizontal and vertical components.
Horizontal component is radial.
\[
F_{\text{radial}}
= T\sin\theta
\]
\[
F_{\text{radial}}
= m\,a_{\text{radial}}= \frac{mv^2}{r}
\]
\[
T\sin\theta
= \frac{mv^2}{r}
\]
\[
T
= \frac{m v^{2}}{r\sin\theta}
= \frac{mr\,\omega^{2}}{\sin\theta}
\]
Vertical component:
has no acceleration, so is equal in magnitude but opposite in direction to mg.
\[F_\text{vertical}
= T\cos\theta
\]
\[
F_{\text{vertical}} + mg = 0
\]
\[
|T\cos\theta| = mg
\]
\[
|T| = \frac{mg}{\cos\theta}
\]
\[
T
= \frac{mr\,\omega^{2}}{\sin\theta}
\]
\[
\omega^{2}
= \frac{T\sin\theta}{mr}
\]
\[
\omega
= \sqrt{
\frac{T\sin\theta}{mr}
}
\]
A mass of 300g is attached to a light inextensible string of length 75 cm which is attached to point A. The mass is set to rotate with a constant speed in a circular motion of radius 60 cm about a point which is vertically below A.
Calculate the tension,T, in the string and the angular speed ,ω.
\[
\sin\theta
= \frac{0.6}{0.75}
\]
\[
\theta
= \sin^{-1}\!\left(\frac{0.6}{0.75}\right)
\]
\[
\theta = 53.1301^\circ
\]
\[
\theta = 53.1^\circ \;\text{(1dp)}
\]
\[
|T|
= \frac{mg}{\cos\theta}
\]
\[
= \frac{0.3 \times 9.81}{\cos 53.1^\circ}
\]
\[
= 4.901
\]
\[
= 4.9\,N \;\text{(1dp)}
\]
\[
\omega
= \sqrt{
\frac{T\sin\theta}{mr}
}
\]
\[
= \sqrt{
\frac{4.9 \times (0.6/0.75)}
{0.3 \times 0.6}
}
\]
\[
= \sqrt{
\frac{196}{9}
}
\]
\[
= \sqrt{21.777}
\]
\[
= 4.67\;\text{rads}\;(\text{2dp})
\]
Newton’s Inverse Square Law
Sir Isaac Newton realised that there was an inverse square relationship between the gravitational effect of a body and the distance from it.
\[
F \propto \frac{1}{r^{2}}
\]
\[
F = \frac{k}{r^{2}}
\]
The force between two masses separated by a distance, r, is
\(
F = G\frac{m_1m_2}{r^2}
\) ,
where G is the gravitational constant,
\[
6.674 \times 10^{-11}\;
m^{3}\,kg^{-1}\,s^{-2}
\]
\[
= 6.674 \times 10^{-11}\;
\frac{m^{3}}{kg\,s^{2}}
\]
\[
= 6.674 \times 10^{-11}\;
\frac{N\,m^{2}}{kg^{2}}
\]
Gravitation Near Earth
Let M = mass of the Earth
The gravitational force from Earth on a particle of mass m , located a distance of r metres from the Earth's centre is given by
\[
F = \frac{GMm}{r^2}
\]
\[
g = \frac{GM}{r^2}
\]
The particle will accelerate towards Earth with acceleration due to gravity, called gravitational acceleration.
so
\[
F = \frac{GMm}{r^{2}}
\]
\[
m\,a_E
= \frac{GMm}{r^{2}}
\]
\[
a_E
= \frac{GM}{r^{2}}
\]
\[
r^{2}
= \frac{GM}{a_E}
\]
\[
r
= \sqrt{
\frac{GM}{a_E}
}
\]
Since the force is attractive, it points backwards towards the source, so a can be replaced by g.
A probe is in a circular orbit above Venus, and experiences acceleration due to gravity at 6.6 m/s2.
Venus has a radius of 6052 km and has acceleration due to gravity on the surface of 8.867 m/s2.
How high above the surface of Venus is the probe ?
On the surface of Venus
\[
F = \frac{k}{r^{2}}
\]
\[
m\,g_V
= \frac{k}{(6052000)^{2}}
\]
\[
k
= (6052000)^{2}\, m\,g_V
\]
At an altitude above Venus,
\[
F = \frac{k}{r^{2}}
\]
\[
m\,g_{vA}
= \frac{k}{r^{2}}
\]
\[
k
= r^{2}\,m\,g_{VA}
\]
Equating for k
\[
k
= r^{2}\,m\,g_{VA}
\]
\[
k
= (6052000)^{2}\, m\,g_{V}
\]
\[
r^{2}\,m\,g_{VA}
= (6052000)^{2}\, m\,g_{V}
\]
\[
r^{2}
= \frac{
(6052000)^{2}\, m\,g_{V}
}{
m\,g_{VA}
}
\]
\[
r
= \sqrt{
\frac{
(6052000)^{2}\, g_{V}
}{
g_{VA}
}
}
\]
\[
r
= \sqrt{
\frac{
(6052000)^{2} \times 8.867
}{
6.6
}
}
\]
\[
r = 7014800.202
\]
\[
7014800.202 - 6052000 = 962800.62
\]
\[
\text{The probe is approximately 962.8 km above the surface of Venus}
\]
