Question

A motorcyclist wants to drive on the vertical surface of a wooden well of radius $5m$, with a minimum speed of $5\sqrt{5}$. The minimum value of coefficient of friction between the tyres and the wall of the wood must be: (Take $g=10m/s^2$)

• A. $0.10$
• B. $0.20$
• C. $0.30$
• D. $0.40$

Answer 1

The correct option is D

$0.40$

Step 1: Given

The radius of the well: $r=5m$

Minimum speed: $v=5\sqrt{5}$

Acceleration due to gravity: $g=10m/s^2$

Step 2: Formula Used

Frictional Force: $F=mg$

Centripetal force: $N=\frac{m{v}^2}{r}$

Coefficient of friction: $\mu =\frac{F}{N}$

Step 3: Calculating coefficient of friction

Substitute $F=mg$ and $N=\frac{m{v}^2}{r}$ in $\mu =\frac{F}{N}$.

$$\begin{gathered} \mu =\frac{F}{N} \\ =\frac{mg}{{\displaystyle \frac{m{v}^2}{r}}} \\ =\frac{gr}{v^2} \end{gathered}$$

Substitute the values in the obtained equation to calculate the coefficient

$$\begin{gathered} \mu =\frac{10\times 5}{{\left(5\sqrt{5}\right)}^2} \\ =\frac{50}{125} \\ =\frac{2}{5} \\ =0.40 \end{gathered}$$

Hence, option D is correct.