Question 172: Work, Energy and Power & Circular Motion

Question

A curved in a level road has a radius 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be ___________ m/s.

Accepted Answer

For a car to move safely around a curved road without skidding, the centripetal force needed is provided by the frictional force between the tires and the road. The relationship governing this scenario can be expressed as:
$$  f_{	ext{friction}} = f_{	ext{centripetal}} $$
The frictional force is given by:
$$ f_{	ext{friction}} = \mu \cdot m \cdot g $$
And the centripetal force is given by:
$$ f_{	ext{centripetal}} = \frac{m \cdot v^2}{r} $$
Where:
$\mu$ is the coefficient of friction,
$m$ is the mass of the car,
$g$ is the acceleration due to gravity (approximately $9.8 \, 	ext{m/s}^2$),
$v$ is the speed of the car,
$r$ is the radius of the curve.
Since the frictional force equals the centripetal force, we have:
$$ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} $$
Cancelling $m$ from both sides (assuming $m 
eq 0$) gives:
$$ \mu \cdot g = \frac{v^2}{r} $$
Rewriting for $v$ gives the speed:
$$ v = \sqrt{\mu \cdot g \cdot r} $$
For the first scenario with $r = 75 \, 	ext{m}$ and $v = 30 \, 	ext{m/s}$:
$$ v_1^2 = \mu \cdot g \cdot r_1 $$
Thus,
$$ \mu \cdot g = \frac{v_1^2}{r_1} = \frac{30^2}{75} = \frac{900}{75} = 12 $$
For the second scenario with $r_2 = 48 \, 	ext{m}$, the new speed $v_2$ is:
$$ v_2 = \sqrt{\mu \cdot g \cdot r_2} = \sqrt{12 	imes 48} $$
Calculating inside the square root:
$$ 12 	imes 48 = 576 $$
So,
$$ v_2 = \sqrt{576} = 24 $$
Hence, the maximum allowed speed for a curve with a radius of $48 \, 	ext{m}$ is
24 m/s
.

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type