Question

A stunt driver of mass $60\text{kg}$ rides a bike inside a hollow metallic sphere of radius $45\text{m}$. Assume the bike moves in the central plane of the sphere. If the coefficient of static friction between the wheels of the bike and the metal surface is $0.8$, what must be the minimum speed of the driver so that he does not crash? Take $g=10{\text{m/s}}^2$.

• A. $23.72\text{m/s}$
• B. $40\text{m/s}$
• C. $25.72\text{m/s}$
• D. $50\text{m/s}$

Answer 1

The correct option is A $23.72\text{m/s}$

Since the motion in the central plane of the sphere, the normal force on the bike will be horizontal. In this case, driver will tend to slip downwards, hence friction will act upwards to support the weight of driver.



Friction will act at its maximum value to prevent slipping, because rider is moving at minimum speed.

From the FBD:
$N=\frac{m{v}^2}{r}$
(applying equation of circular dynamics)
and $f=mg$
(in vertical direction, forces should be balanced)

Since we are asked for the minimum speed of the driver for no slipping, we must consider the limiting value of friction.
i.e $N=\frac{m{v}_{min}^2}{r}-\left(1\right)$ and
$f_{max}=\mu N=mg-\left(2\right)$

Combining (1) and (2),
$\mu \frac{m{v}_{min}^2}{r}=mg$
$\Rightarrow v_{min}=\sqrt{\frac{gr}{\mu }}=\sqrt{\frac{10\times 45}{0.8}}$
$=23.72\text{m/s}$