Circular Motion
9.0 Bending of a cyclist or motorcyclist while taking turn
9.0 Bending of a cyclist or motorcyclist while taking turn
The motorcycle is running on a circular track of radius $r$ with speed $v$.
We will examine two cases,
When motorcycle is vertical
When motorcycle is inclined
Case 1: When motorcycle is vertical
$$\begin{equation} \begin{aligned} f = \frac{{m{v^2}}}{r} \\ N = mg \\\end{aligned} \end{equation} $$
Moment about $CG$, $${M_{CG}} = fH$$
So, we can see that the moment can turn the motorcycle outwards.
Case 2: When motorcycle is inclined
$$\begin{equation} \begin{aligned} f = \frac{{m{v^2}}}{r} \\ N = mg \\\end{aligned} \end{equation} $$
Moment about $CG$, $${M_{CG}} = fh\cos \theta - Nh\sin \theta $$
Therefore for no turning of motorcycle outwards, $$\begin{equation} \begin{aligned} {M_{CG}} = 0 \\ \tan \theta = \frac{f}{N} \\ \tan \theta = \frac{{{v^2}}}{{rg}} \\\end{aligned} \end{equation} $$
So, the inclined motorcycle is safe while taking circular turn.
Question 10. A motorcycle weighing $1200\ kg$ is running with a speed of $108\ km/h$ in a circular track of radius $100\ m$. Find the frictional force acting between the tires. Also, what is the angle of inclination for safe turning?
Solution: Speed of motorcycle, $$\begin{equation} \begin{aligned} v = 108\;km/h \\ v = 30\;m/s \\\end{aligned} \end{equation} $$
Radius of circular track, $$r = 1000\;m$$
Mass of the motorcycle, $$m = 1200\;kg$$
As we know, $$\begin{equation} \begin{aligned} f = \frac{{m{v^2}}}{r} \\ f = \frac{{1200 \times {{30}^2}}}{{100}} \\ f = 10800\;N \\\end{aligned} \end{equation} $$ and $$\begin{equation} \begin{aligned} \tan \theta = \frac{{{v^2}}}{{rg}} \\ \tan \theta = \frac{{{{30}^2}}}{{100 \times 10}} \\ \tan \theta = 0.9 \\\end{aligned} \end{equation} $$
