Racetrack Banking Angle Physics
The problem asks for the banking angle ($\theta$) of a circular racetrack of radius $r = 50\text{ m}$ where a car traveling at a speed $v = 10\text{ ms}^{-1}$ experiences minimum wear and tear on its tires. Minimum wear typically occurs when the centripetal force is provided solely by the horizontal component of the normal force, meaning friction is not needed.
Centripetal Force Condition
For minimum tire wear on a banked curve, the net horizontal force towards the center of the circle must equal the centripetal force required for the circular motion. This condition can be derived by considering the forces acting on the car:
- Vertical forces: The vertical component of the normal force ($N$) balances the gravitational force ($mg$). $N \cos(\theta) = mg \quad (1)$
- Horizontal forces: The horizontal component of the normal force provides the necessary centripetal force ($\frac{mv^2}{r}$). $N \sin(\theta) = \frac{mv^2}{r} \quad (2)$
Calculating Banking Angle
To find the angle $\theta$, divide equation (2) by equation (1):
$ \frac{N \sin(\theta)}{N \cos(\theta)} = \frac{\frac{mv^2}{r}}{mg} $
$ \tan(\theta) = \frac{v^2}{rg} $
Now, substitute the given values:
- $v = 10\text{ ms}^{-1}$
- $r = 50\text{ m}$
- $g = 10\text{ ms}^{-2}$
Calculation:
$ \tan(\theta) = \frac{(10\text{ ms}^{-1})^2}{(50\text{ m})(10\text{ ms}^{-2})} $ $ \tan(\theta) = \frac{100 \text{ m}^2/\text{s}^2}{500 \text{ m}^2/\text{s}^2} $ $ \tan(\theta) = \frac{1}{5} $Therefore, the banking angle is:
$ \theta = \tan^{-1}\left(\frac{1}{5}\right) $Final Answer
The value of $\theta$ is $\tan^{-1}\left(\frac{1}{5}\right)$.
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