Calculating Maximum Speed for Cyclist Turn in Velodrome

The question asks for the maximum speed a cyclist can take a turn in a velodrome without leaning inwards, given the radius of the velodrome and the coefficient of friction between the tires and the track. When a cyclist takes a turn on a flat surface without leaning, the necessary centripetal force required to maintain circular motion is provided solely by the static friction force between the tires and the surface.

Understanding Forces in a Non-Leaning Turn

Consider the cyclist moving in a circular path of radius $r$ on a flat horizontal surface without leaning. The forces acting on the cyclist are:

• Gravitational force ($mg$) acting vertically downwards.
• Normal force ($N$) acting vertically upwards from the surface.
• Static friction force ($f_s$) acting horizontally towards the center of the circular path.

Since there is no vertical acceleration, the normal force balances the gravitational force:

$N = mg$

The static friction force provides the centripetal force required for the circular motion. The magnitude of the centripetal force required for a cyclist moving at speed $v$ in a circle of radius $r$ is given by:

$F_c = \frac{mv^2}{r}$

For the cyclist to successfully take the turn, the static friction force must be equal to the required centripetal force. The maximum static friction force available is given by:

$f_{s,max} = \mu_s N = \mu_s mg$

To find the maximum speed ($v_{max}$) with which the cyclist can take the turn without slipping (and without leaning), the required centripetal force at this speed must be equal to the maximum static friction force:

$\frac{mv_{max}^2}{r} = f_{s,max}$

$\frac{mv_{max}^2}{r} = \mu_s mg$

Deriving the Maximum Speed Formula

We can cancel the mass ($m$) from both sides of the equation:

$\frac{v_{max}^2}{r} = \mu_s g$

Now, solve for $v_{max}$:

$v_{max}^2 = \mu_s g r$

$v_{max} = \sqrt{\mu_s g r}$

This formula gives the maximum speed a vehicle or person can take a turn on a flat surface relying only on friction.

Calculating the Maximum Speed

Given values:

• Radius of the velodrome, $r = 80$ m
• Coefficient of friction, $\mu_s = 0.25$
• Acceleration due to gravity, $g \approx 9.8$ m/s²

Substitute these values into the formula:

$v_{max} = \sqrt{0.25 \times 9.8 \times 80}$

$v_{max} = \sqrt{0.25 \times 784}$

$v_{max} = \sqrt{196}$

$v_{max} = 14$ m/s

Therefore, the maximum speed with which the cyclist can take a turn without leaning inwards is 14 m/s.

Conclusion

Based on the calculation, the maximum speed is 14 m/s. This matches one of the provided options.

Revision Table: Circular Motion and Friction

Additional Information: Banking and Leaning

In real-world scenarios like velodromes or race tracks, turns are often banked. Banking involves tilting the track surface inwards. This banking allows a component of the normal force to contribute to the required centripetal force, in addition to friction. This allows vehicles or cyclists to take turns at higher speeds without relying solely on friction, which is especially important when friction is reduced (e.g., wet surface).

For cyclists, leaning inwards while taking a turn on a flat surface also helps. Leaning allows the total reaction force from the ground (combination of normal force and friction) to provide the necessary inward force (centripetal force). The angle of lean changes the direction of the normal force and friction force relative to the vertical and horizontal axes, enabling a larger inward component without exceeding the maximum static friction limit for a given speed. However, the question specifically asks about the speed without *leaning inwards*, meaning the scenario is simplified to rely only on the horizontal friction force on a conceptually flat surface.