U3AOS1 Topic 7: Banked Circular Motion

Banked circular motion is similar to normal circle motion. However, the only difference is that the object is travelling in a circle while it is on a slope. So it is a combination of the inclined planes and circular motions topics.


In this situation, the x component of the normal force points toward the center of the circle and is responsible for the net acceleration toward the center. However, there is a speed at which the object will not be accelerated toward the center.


This is called the design speed and is because the friction forces cancel out the centripetal forces. It is calculated through the following formula:

\[ \displaystyle \Large v= \sqrt{rg \times \tan(\theta)} \]


Use the simulation below to grasp the concept.



Created with GeoGebra®, by Mark Willis, Link


Now have a read through the following examples to better understand how to attempt banking questions. 

Example 1

A car with a mass of $ 1200 $ kg is traveling on a banked curve with a radius of $ 100 $ meters. The banking angle (θ) is $ 30 $ degrees. Calculate the design speed at which the car can safely navigate the curve without experiencing any acceleration towards the center. Assume the coefficient of friction between the tires and the road is $ 0.2 $.

First, make note of all the data provided in the question.
We are asked to calculate  $ ( V_{ \text{design} } $: 
\[ m_{\text{car}} = 1200 \text{kg} \] 
\[  r = 100 \text{m} \]  
\[ \theta = 30 \degree \] 
\[ a_{\text{centripetal}} = 0 \text{m/} \text{s}^{2} \] 
\[ g=9.81 \text{ms}^{-2} \] 
The formula for design speed is: $ V = rg \tan( \theta )​ $

Now, plug in the known values to calculate the design speed:
\[ v_{\text{design}}= \sqrt{100 \times 9.81 \times \tan(30)} \]
\[ v_{\text{design}}= 23.79875 \]

So the design speed is \( 23.8 m/s \).

Example 2

A cyclist of mass 50 kg is riding around a velodrome with a bank angle of 30 degrees to the horizontal. The radius of the track is 25m.
(a) Determine the speed of the cyclist if he experiences no sideways force on his bike as he rides the around the velodrome. 
(b)     Calculate the normal force acting on the cyclist. How does this compare to if he was riding on a flat track instead?

As always make note of the information provided to you in the question.
\[ m= 30kg \] 
\[ \theta = 30 \text{ degrees} \] 
\[ r= 25m \] 
\[ g= 9.81 m/s^2 \] 
You know what's a good idea? Draw a diagram.
(a)
As the question states that the cyclist experiences no sideways force, he must be travelling at design speed. This means that the cyclist's mass of 50kg has no effect on Vdesign 
So,
\[ V_{\text{design}}= \sqrt{rg \tan(\theta)}\]
\[ V_{\text{design}}= \sqrt{25 \times 9.81 \times \tan(30)}\]
\[ V_{\text{design}}=11.89937\]

 
So the design speed is 11.9 m/s.














So the normal force is \(424.8 N \) perpendicular to the surface of the velodrome track. 

Now how does this compare to riding on a flat track?
Well if he was riding on a flat surface his normal force would simply be equal to his weight force (m*g). In this case, gravity is the only vertical force acting on the cyclist hence the normal force acts in the opposite direction to prevent the cyclist sinking into the ground! In the velodrome however, the normal force is not the only force acting to balance the cyclist and keep him moving around the velodrome without falling (or sinking!), but there is also a centripetal force acting towards the center of the velodrome circle. So he would actually feel lighter than he would if he was on a flat track. 



Back: U3AOS1 Topic 6: Inclined Planes Next: U3AOS1 Topic 8: Vertical Circular Motion