Physics A Level
Chapter 16: Circular motion 16.6 Calculating acceleration and force
Physics A Level
Chapter 16: Circular motion 16.6 Calculating acceleration and force
If we spin a bung around in a circle (Figure 16.7), we get a feeling for the factors that determine the centripetal force F required to keep it in its circular orbit. The greater the mass m of the bung and the greater its speed v, the greater is the force F that is required. However, if the radius r of the circle is increased, F is smaller.
Now we will deduce an expression for the centripetal acceleration of an object moving around a circle with a constant speed.
Figure 16.11 shows a particle moving round a circle. In time $\Delta t$ it moves through an angle $\Delta \theta $ from A to B.
Its speed remains constant but its velocity changes by $\Delta v$, as shown in the vector diagram. Since the narrow angle in this triangle is also $\Delta \theta $, we can say that:
$\Delta \theta = \frac{{\Delta v}}{v}$
Dividing both sides of this equation by $\Delta t$ and rearranging gives:
$\frac{{\Delta v}}{{\Delta t}} = \frac{{v\Delta \theta }}{{\Delta t}}$
The quantity on the left is $\frac{{\Delta v}}{t} = a$, the particle’s acceleration.
The quantity on the right is $\frac{{\Delta \theta }}{{\Delta t}} = \omega $, the angular velocity.
Substituting for these gives:
$a = v\omega $
Using $v = \omega r$, we can eliminate $\omega $ from this equation:
$a = \frac{{{v^2}}}{r}$
where a is the centripetal acceleration, v is the speed and r is the radius of the circle.
Question
13) Show that an alternative equation for the centripetal acceleration is $a = {\omega ^2}r$.
Newton’s second law of motion
Now that we have an equation for centripetal acceleration, we can use Newton’s second law of motion to deduce an equation for centripetal force. If we write this law as $F = ma$, we find:
$\begin{array}{l}
centripetal\,force\,F = \frac{{m{v^2}}}{r}\\
= mr{\omega ^2}
\end{array}$
Remembering that an object accelerates in the direction of the resultant force on it, it follows that both F and a are in the same direction, towards the centre of the circle.
Questions
14) Calculate how long it would take a ball to orbit the Earth once, just above the surface, at a speed of $7920\,m\,{s^{ - 1}}$. (The radius of the Earth is $6400 km$.)
15) A stone of mass $0.20 kg$ is whirled round on the end of a string in a vertical circle of radius $30 cm$.
The string will break when the tension in it exceeds $8.0 N$. Calculate the maximum speed at which the stone can be whirled without the string breaking.
Figure 16.12: The view from the International Space Station, orbiting the Earth over Australia.
16) The International Space Station (Figure 16.12) has a mass of 350 tonnes, and orbits the Earth at an average height of $340 km$ where the gravitational acceleration is $8.8\,m\,{s^{ - 2}}$. The radius of the Earth is $6400 km$. Calculate:
a: the centripetal force on the space station
b: the speed at which it orbits
c: the time taken for each orbit
d: the number of times it orbits the Earth each day.
17) An toy truck of mass $0.40 kg$ travels round a horizontal circular track of radius $0.50 m$. It makes three complete revolutions every 10 seconds.
Calculate:
a: its speed
b: its centripetal acceleration
c: the centripetal force.
18) Mars orbits the Sun once every 687 days at a distance of $2.3 \times {10^{11}}\,m$. The mass of Mars is $6.4 \times {10^{23}}\,kg$.
Calculate:
a: the average speed in metres per second
b: its centripetal acceleration
c: the gravitational force exerted on Mars by the Sun.
Calculating orbital speed
We can use the force equation to calculate the speed that an object must have to orbit the Earth under gravity, as in Newton’s thought experiment. The necessary centripetal force $\frac{{m{v^2}}}{r}$ is provided by the Earth’s gravitational pull mg.
Hence:
$\begin{array}{l}
mg = \frac{{m{v^2}}}{r}\\
g = \frac{{{v^2}}}{r}
\end{array}$
where $g = 9.81\,m\,{s^{ - 2}}$ is the acceleration of free fall close to the Earth’s surface. The radius of its orbit is equal to the Earth’s radius, approximately $6400 km$. Hence, we have:
$\begin{array}{l}
9.81 = \frac{{{v^2}}}{{(6.4 \times {{10}^6})}}\\
{v^2} = 9.81 \times (6.4 \times {10^6})\\
v \approx 7.92 \times {10^3}\,m\,{s^{ - 1}}
\end{array}$
Thus, if you were to throw or hit a ball horizontally at almost $8\,km\,{s^{ - 1}}$, it would go into orbit around the Earth.